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Eureka Math™

Grade 7, Module 3

Teacher Edition

A Story of Ratios®

Module 3: Expressions and Equations

7 G R A D E

Mathematics Curriculum GRADE 7 • MODULE 3

Table of Contents1

Expressions and Equations Module Overview .................................................................................................................................................. 3

Topic A: Use Properties of Operations to Generate Equivalent Expressions (7.EE.A.1, 7.EE.A.2) ...................... 12

Lessons 1–2: Generating Equivalent Expressions ................................................................................... 13

Lessons 3–4: Writing Products as Sums and Sums as Products ............................................................. 48

Lesson 5: Using the Identity and Inverse to Write Equivalent Expressions............................................ 73

Lesson 6: Collecting Rational Number Like Terms .................................................................................. 85

Topic B: Solve Problems Using Expressions, Equations, and Inequalities (7.EE.B.3, 7.EE.B.4, 7.G.B.5) ............. 98

Lesson 7: Understanding Equations ..................................................................................................... 100

Lessons 8–9: Using If-Then Moves in Solving Equations ...................................................................... 113

Lessons 10–11: Angle Problems and Solving Equations ....................................................................... 150

Lesson 12: Properties of Inequalities ................................................................................................... 171

Lesson 13: Inequalities ......................................................................................................................... 184

Lesson 14: Solving Inequalities ............................................................................................................. 193

Lesson 15: Graphing Solutions to Inequalities ..................................................................................... 203

Mid-Module Assessment and Rubric ................................................................................................................ 217 Topics A through B (assessment 2 days, return 1 day, remediation or further applications 2 days)

Topic C: Use Equations and Inequalities to Solve Geometry Problems (7.G.B.4, 7.G.B.6) ............................... 239

Lesson 16: The Most Famous Ratio of All ............................................................................................. 241

Lesson 17: The Area of a Circle ............................................................................................................. 252

Lesson 18: More Problems on Area and Circumference ...................................................................... 262

Lesson 19: Unknown Area Problems on the Coordinate Plane ............................................................ 273

Lesson 20: Composite Area Problems ................................................................................................. 283

Lessons 21–22: Surface Area ................................................................................................................ 295

Lessons 23–24: The Volume of a Right Prism ....................................................................................... 320

1Each lesson is ONE day, and ONE day is considered a 45-minute period.

A STORY OF RATIOS

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7•3 Module Overview

Lessons 25–26: Volume and Surface Area ............................................................................................ 342

End-of-Module Assessment and Rubric ............................................................................................................ 364 Topics A through C (assessment 1 day, return 1 day, remediation or further applications 2 days)

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Grade 7 • Module 3

Expressions and Equations OVERVIEW In Grade 6, students interpreted expressions and equations as they reasoned about one-variable equations (6.EE.A.2). This module consolidates and expands upon students’ understanding of equivalent expressions as they apply the properties of operations (associative, commutative, and distributive) to write expressions in both standard form (by expanding products into sums) and in factored form (by expanding sums into products). They use linear equations to solve unknown angle problems and other problems presented within context to understand that solving algebraic equations is all about the numbers. It is assumed that a number already exists to satisfy the equation and context; we just need to discover it. A number sentence is an equation that is said to be true if both numerical expressions evaluate to the same number; it is said to be false otherwise. Students use the number line to understand the properties of inequality and recognize when to preserve the inequality and when to reverse the inequality when solving problems leading to inequalities. They interpret solutions within the context of problems. Students extend their sixth-grade study of geometric figures and the relationships between them as they apply their work with expressions and equations to solve problems involving area of a circle and composite area in the plane, as well as volume and surface area of right prisms. In this module, students discover the most famous ratio of all, 𝜋𝜋, and begin to appreciate why it has been chosen as the symbol to represent the Grades 6–8 mathematics curriculum, A Story of Ratios.

To begin this module, students will generate equivalent expressions using the fact that addition and multiplication can be done in any order with any grouping and will extend this understanding to subtraction (adding the inverse) and division (multiplying by the multiplicative inverse, also known as the reciprocal) (7.EE.A.1). They extend the properties of operations with numbers (learned in earlier grades) and recognize how the same properties hold true for letters that represent numbers. Knowledge of rational number operations from Module 2 is demonstrated as students collect like terms containing both positive and negative integers.

An area model is used as a tool for students to rewrite products as sums and sums as products and to provide a visual representation leading students to recognize the repeated use of the distributive property in factoring and expanding linear expressions (7.EE.A.1). Students examine situations where more than one form of an expression may be used to represent the same context, and they see how looking at each form can bring a new perspective (and thus deeper understanding) to the problem. Students recognize and use the identity properties and the existence of additive inverses to efficiently write equivalent expressions in standard form, for example, 2𝑥𝑥 + (−2𝑥𝑥) + 3 = 0 + 3 = 3 (7.EE.A.2). By the end of the topic, students have the opportunity to practice knowledge of operations with rational numbers gained in Module 2 (7.NS.A.1, 7.NS.A.2) as they collect like terms with rational number coefficients (7.EE.A.1).

In Topic B, students use linear equations and inequalities to solve problems (7.EE.B.4). They continue to use tape diagrams from earlier grades where they see fit, but will quickly discover that some problems would more reasonably be solved algebraically (as in the case of large numbers). Guiding students to arrive at this

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realization on their own develops the need for algebra. This algebraic approach builds upon work in Grade 6 with equations (6.EE.B.6, 6.EE.B.7) to now include multi-step equations and inequalities containing rational numbers (7.EE.B.3, 7.EE.B.4). Students solve problems involving consecutive numbers; total cost; age comparisons; distance, rate, and time; area and perimeter; and missing angle measures. Solving equations with a variable is all about numbers, and students are challenged with the goal of finding the number that makes the equation true. When given in context, students recognize that a value exists, and it is simply their job to discover what that value is. Even the angles in each diagram have a precise value, which can be checked with a protractor to ensure students that the value they find does indeed create a true number sentence.

In Topic C, students continue work with geometry as they use equations and expressions to study area, perimeter, surface area, and volume. This final topic begins by modeling a circle with a bicycle tire and comparing its perimeter (one rotation of the tire) to the length across (measured with a string) to allow students to discover the most famous ratio of all, pi. Activities in comparing circumference to diameter are staged precisely for students to recognize that this symbol has a distinct value and can be approximated by 227

, or 3.14, to give students an intuitive sense of the relationship that exists. In addition to representing this

value with the 𝜋𝜋 symbol, the fraction and decimal approximations allow for students to continue to practice their work with rational number operations. All problems are crafted in such a way as to allow students to

practice skills in reducing within a problem, such as using 227

for finding circumference with a given diameter

length of 14 cm, and recognize what value would be best to approximate a solution. This understanding allows students to accurately assess work for reasonableness of answers. After discovering and understanding the value of this special ratio, students will continue to use pi as they solve problems of area and circumference (7.G.B.4).

In this topic, students derive the formula for area of a circle by dividing a circle of radius 𝑟𝑟 into pieces of pi and rearranging the pieces so that they are lined up, alternating direction, and form a shape that resembles a

rectangle. This “rectangle” has a length that is 12

the circumference and a width of 𝑟𝑟. Students determine

that the area of this rectangle (reconfigured from a circle of the same area) is the product of its length and its

width: 12𝐶𝐶 ∙ 𝑟𝑟 =

12

2𝜋𝜋𝑟𝑟 ∙ 𝑟𝑟 = 𝜋𝜋𝑟𝑟2 (7.G.B.4). The precise definitions for diameter, circumference, pi, and

circular region or disk will be developed during this topic with significant time being devoted to students’ understanding of each term.

Students build upon their work in Grade 6 with surface area and nets to understand that surface area is simply the sum of the area of the lateral faces and the base(s) (6.G.A.4). In Grade 7, they continue to solve real-life and mathematical problems involving area of two-dimensional shapes and surface area and volume of prisms (e.g., rectangular, triangular), focusing on problems that involve fractional values for length (7.G.B.6). Additional work (examples) with surface area will occur in Module 6 after a formal definition of rectangular pyramid is established.

This module is comprised of 26 lessons; 9 days are reserved for administering the Mid-Module and End-of-Module Assessments, returning the assessments, and remediating or providing further applications of the concepts. The Mid-Module Assessment follows Topic B, and the End-of-Module Assessment follows Topic C.

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Focus Standards Use properties of operations to generate equivalent expressions.

7.EE.A.1 Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.

7.EE.A.2 Understand that rewriting an expression in different forms in a problem context can shed light on the problem and how the quantities in it are related. For example, 𝑎𝑎 + 0.05𝑎𝑎 = 1.05𝑎𝑎 means that “increase by 5%” is the same as “multiply by 1.05.”

Solve real-life and mathematical problems using numerical and algebraic expressions and equations.

7.EE.B.3 Solve multi-step real-life and mathematical problems posed with positive and negative rational numbers in any form (whole numbers, fractions, and decimals), using tools strategically. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate; and assess the reasonableness of answers using mental computation and estimation strategies. For example: If a woman making $25 an hour gets a 10% raise, she will make an additional 1/10 of her salary an hour, or $2.50, for a new salary of $27.50. If you want to place a towel bar 9 3/4 inches long in the center of a door that is 27 1/2 inches wide, you will need to place the bar about 9 inches from each edge; this estimate can be used as a check on the exact computation.

7.EE.B.4 Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities.

a. Solve word problems leading to equations of the form 𝑝𝑝𝑥𝑥 + 𝑞𝑞 = 𝑟𝑟 and 𝑝𝑝(𝑥𝑥 + 𝑞𝑞) = 𝑟𝑟, where 𝑝𝑝, 𝑞𝑞, and 𝑟𝑟 are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach. For example, the perimeter of a rectangle is 54 cm. Its length is 6 cm. What is its width?

b. Solve word problems leading to inequalities of the form 𝑝𝑝𝑥𝑥 + 𝑞𝑞 > 𝑟𝑟 or 𝑝𝑝𝑥𝑥 + 𝑞𝑞 < 𝑟𝑟, where 𝑝𝑝, 𝑞𝑞, and 𝑟𝑟 are specific rational numbers. Graph the solution set of the inequality and interpret it in the context of the problem. For example: As a salesperson, you are paid $50 per week plus $3 per sale. This week you want your pay to be at least $100. Write an inequality for the number of sales you need to make, and describe the solutions.

Solve real-life and mathematical problems involving angle measure, area, surface area, and volume.

7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle.

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7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure.

7.G.B.6 Solve real-world and mathematical problems involving area, volume and surface area of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms.

Foundational Standards Understand and apply properties of operations and the relationship between addition and subtraction.

1.OA.B.3 Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)

Understand properties of multiplication and the relationship between multiplication and division.

3.OA.B.5 Apply properties of operations as strategies to multiply and divide.2 Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.)

Geometric measurement: understand concepts of angle and measure angles.

4.MD.C.5 Recognize angles as geometric shapes that are formed wherever two rays share a common endpoint, and understand concepts of angle measurement:

a. An angle is measured with reference to a circle with its center at the common endpoint of the rays, by considering the fraction of the circular arc between the points where the two rays intersect the circle. An angle that turns through 1/360 of a circle is called a “one-degree angle,” and can be used to measure angles.

b. An angle that turns through 𝑛𝑛 one-degree angles is said to have an angle measure of 𝑛𝑛 degrees.

4.MD.C.6 Measure angles in whole-number degrees using a protractor. Sketch angles of specified measure.

2Students need not use formal terms for these properties.

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4.MD.C.7 Recognize angle measure as additive. When an angle is decomposed into non-overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Solve addition and subtraction problems to find unknown angles on a diagram in real world and mathematical problems, e.g., by using an equation with a symbol for the unknown angle measure.

Apply and extend previous understandings of arithmetic to algebraic expressions.

6.EE.A.3 Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3(2 + 𝑥𝑥) to produce the equivalent expression 6 + 3𝑥𝑥; apply the distributive property to the expression 24𝑥𝑥 + 18𝑦𝑦 to produce the equivalent expression 6(4𝑥𝑥 + 3𝑦𝑦); apply properties of operations to 𝑦𝑦 + 𝑦𝑦 + 𝑦𝑦 to produce the equivalent expression 3𝑦𝑦.

6.EE.A.4 Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions 𝑦𝑦 + 𝑦𝑦 + 𝑦𝑦 and 3𝑦𝑦 are equivalent because they name the same number regardless of which number 𝑦𝑦 stands for.

Reason about and solve one-variable equations and inequalities.

6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.

6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations in the form 𝑥𝑥 + 𝑝𝑝 = 𝑞𝑞 and 𝑝𝑝𝑥𝑥 = 𝑞𝑞 for cases in which 𝑝𝑝, 𝑞𝑞, and 𝑥𝑥 are all nonnegative rational numbers.

6.EE.B.8 Write an inequality of the form 𝑥𝑥 > 𝑐𝑐 or 𝑥𝑥 < 𝑐𝑐 to represent a constraint or condition in a real-world mathematical problem. Recognize that inequalities of the form 𝑥𝑥 > 𝑐𝑐 or 𝑥𝑥 < 𝑐𝑐 have infinitely many solutions; represent solutions of such inequalities on number line diagrams.

Solve real-world and mathematical problems involving area, surface area, and volume.

6.G.A.1 Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.

6.G.A.2 Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas 𝑉𝑉 = 𝑙𝑙𝑙𝑙ℎ and 𝑉𝑉 = 𝑏𝑏ℎ to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems.

6.G.A.4 Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.

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Apply and extend previous understandings of operations with fractions to add, subtract, multiply, and divide rational numbers.

7.NS.A.1 Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram.

a. Describe situations in which opposite quantities combine to make 0. For example, a hydrogen atom has 0 charge because its two constituents are oppositely charged.

b. Understand 𝑝𝑝 + 𝑞𝑞 as the number located a distance |𝑞𝑞| from 𝑝𝑝, in the positive or negative direction depending on whether 𝑞𝑞 is positive or negative. Show that a number and its opposite have a sum of 0 (are additive inverses). Interpret sums of rational numbers by describing real-world contexts.

c. Understand subtraction of rational numbers as adding the additive inverse, 𝑝𝑝 − 𝑞𝑞 = 𝑝𝑝 + (−𝑞𝑞). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts.

d. Apply properties of operations as strategies to add and subtract rational numbers.

7.NS.A.2 Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.

a. Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as (−1)(−1) = 1 and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts.

b. Understand that integers can be divided, provided that the divisor is not zero, and every quotient of integers (with non-zero divisor) is a rational number. If 𝑝𝑝 and 𝑞𝑞 are integers, then −(𝑝𝑝/𝑞𝑞) = (−𝑝𝑝)/𝑞𝑞 = 𝑝𝑝/(−𝑞𝑞). Interpret quotients of rational numbers by describing real-world contexts.

c. Apply properties of operations as strategies to multiply and divide rational numbers.

d. Convert a rational number to a decimal using long division; know that the decimal form of a rational number terminates in 0s or eventually repeats.

Focus Standards for Mathematical Practice MP.2 Reason abstractly and quantitatively. Students make sense of how quantities are related

within a given context and formulate algebraic equations to represent this relationship. They use the properties of operations to manipulate the symbols that are used in place of numbers, in particular, pi. In doing so, students reflect upon each step in solving and recognize that these properties hold true since the variable is really just holding the place for a number. Students analyze solutions and connect back to ensure reasonableness within context.

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MP.4 Model with mathematics. Throughout the module, students use equations and inequalities as models to solve mathematical and real-world problems. In discovering the relationship between circumference and diameter in a circle, they will use real objects to analyze the relationship and draw conclusions. Students test conclusions with a variety of objects to see if the results hold true, possibly improving the model if it has not served its purpose.

MP.6 Attend to precision. Students are precise in defining variables. They understand that a variable represents one number. They use appropriate vocabulary and terminology when communicating about expressions, equations, and inequalities. They use the definition of equation from Grade 6 to understand how to use the equal sign consistently and appropriately. Circles and related notions about circles are precisely defined in this module.

MP.7 Look for and make use of structure. Students recognize the repeated use of the distributive property as they write equivalent expressions. Students recognize how equations leading to the form 𝑝𝑝𝑥𝑥 + 𝑞𝑞 = 𝑟𝑟 and 𝑝𝑝(𝑥𝑥 + 𝑞𝑞) = 𝑟𝑟 are useful in solving a variety of problems. They see patterns in the way that these equations are solved. Students apply this structure as they understand the similarities and differences in how an inequality of the type 𝑝𝑝𝑥𝑥 + 𝑞𝑞 > 𝑟𝑟 or 𝑝𝑝𝑥𝑥 + 𝑞𝑞 < 𝑟𝑟 is solved.

MP.8 Look for and express regularity in repeated reasoning. Students use area models to write products as sums and sums as products and recognize how this model is a way to organize results from repeated use of the distributive property. As students work to solve problems, they maintain oversight of the process, while attending to the details. They continually evaluate the reasonableness of solutions as they are represented in contexts that allow for students to know that they found the intended value for a given variable. As they solve problems involving pi, they notice how a problem may be reduced by using a given estimate for pi to make calculations more efficient.

Terminology New or Recently Introduced Terms

An Expression in Expanded Form (description) (An expression that is written as sums (and/or differences) of products whose factors are numbers, variables, or variables raised to whole number powers is said to be in expanded form. A single number, variable, or a single product of numbers and/or variables is also considered to be in expanded form.)

An Expression in Factored Form (description) (An expression that is a product of two or more expressions is said to be in factored form.)

An Expression in Standard Form (description) (An expression that is in expanded form where all like terms have been collected is said to be in standard form.)

Circle (Given a point 𝑂𝑂 in the plane and a number 𝑟𝑟 > 0, the circle with center 𝑂𝑂 and radius 𝑟𝑟 is the set of all points in the plane whose distance from the point 𝑂𝑂 is equal to 𝑟𝑟.)

Circular Region or Disk (Given a point 𝐶𝐶 in the plane and a number 𝑟𝑟 > 0, the circular region (or disk) with center 𝐶𝐶 and radius 𝑟𝑟 is the set of all points in the plane whose distance from the point 𝐶𝐶 is less than or equal to 𝑟𝑟.)

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Circumference (The circumference of a circle is the distance around the circle.)3 Coefficient of a Term (The coefficient of a term is the number found by multiplying all of the number

factors in a term together.) Diameter of a Circle (The diameter of a circle is the length of any segment that passes through the

center of a circle whose endpoints lie on the circle. If 𝑟𝑟 is the radius of a circle, then the diameter is 2𝑟𝑟.)

Interior of a Circle (The interior of a circle with center 𝐶𝐶 and radius 𝑟𝑟 is the set of all points in the plane whose distance from the point 𝐶𝐶 is less than 𝑟𝑟.)

Pi (The number pi, denoted 𝜋𝜋, is the value of the ratio given by the circumference to the diameter, that is, 𝜋𝜋 = (circumference)/(diameter).)

Term (description) (Each summand of an expression in expanded form is called a term.)

Familiar Terms and Symbols4

Adjacent Angles Cube Distribute Equation Equivalent Expressions Expression (middle school description) Factor Figure Identity Inequality Length of a Segment Linear Expression Measure of an Angle Number Sentence Numerical Expression (middle school description) Properties of Operations (distributive, commutative, associative) Right Rectangular Prism Segment Square Surface of a Prism Term Triangle

3“Distance around a circular arc” is taken as an undefined term in G-CO.1. 4These are terms and symbols students have seen previously.

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True or False Number Sentence Truth Values of a Number Sentence Value of a Numerical Expression Variable (middle school description) Vertical Angles

Suggested Tools and Representations Area Model Coordinate Plane Equations and Inequalities Expressions Geometric Figures Nets for Three-Dimensional Figures Number Line Protractor Tape Diagram

Assessment Summary Assessment Type Administered Format Standards Addressed

Mid-Module Assessment Task After Topic B Constructed response with rubric

7.EE.A.1, 7.EE.A.2, 7.EE.B.3, 7.EE.B.4, 7.G.B.5

End-of-Module Assessment Task After Topic C Constructed response with rubric 7.EE.A.1, 7.EE.A.2,

7.G.B.4, 7.G.B.5, 7.G.B.6

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Topic A: Use Properties of Operations to Generate Equivalent Expressions

GRADE 7 • MODULE 3

7 G R A D E

Mathematics Curriculum Topic A

Use Properties of Operations to Generate Equivalent Expressions

7.EE.A.1, 7.EE.A.2

Focus Standards: 7.EE.A.1 Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.

7.EE.A.2 Understand that rewriting an expression in different forms in a problem context can shed light on the problem and how the quantities in it are related. For example, 𝑎𝑎 + 0.05𝑎𝑎 = 1.05𝑎𝑎 means that “increase by 5%” is the same as “multiply by 1.05.”

Instructional Days: 6

Lessons 1–2: Generating Equivalent Expressions (P)1

Lessons 3–4: Writing Products as Sums and Sums as Products (P)

Lesson 5: Using the Identity and Inverse to Write Equivalent Expressions (P)

Lesson 6: Collecting Rational Number Like Terms (P)

In Lesson 1 of Topic A, students write equivalent expressions by finding sums and differences extending the any order (commutative property) and any grouping (associative property) to collect like terms and rewrite algebraic expressions in standard form (7.EE.A.1). In Lesson 2, students rewrite products in standard form by applying the commutative property to rearrange like items (numeric coefficients, like variables) next to each other and rewrite division as multiplying by the multiplicative inverse. Lessons 3 and 4 have students using a rectangular array and the distributive property as they first multiply one term by a sum of two or more terms to expand a product to a sum, and then reverse the process to rewrite the sum as a product of the GCF and a remaining factor. Students model real-world problems with expressions and see how writing in one form versus another helps them to understand how the quantities are related (7.EE.A.2). In Lesson 5, students recognize that detecting inverses and the identity properties of 0 for addition and 1 for multiplication allows for ease in rewriting equivalent expressions. This topic culminates with Lesson 6 with students applying repeated use of the distributive property as they collect like terms containing fractional coefficients to rewrite rational number expressions.

1Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson

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Lesson 1: Generating Equivalent Expressions

7•3 Lesson 1


Student Outcomes

Students generate equivalent expressions using the fact that addition and multiplication can be done in any order (commutative property) and any grouping (associative property).

Students recognize how any order, any grouping can be applied in a subtraction problem by using additive inverse relationships (adding the opposite) to form a sum and likewise with division problems by using the multiplicative inverse relationships (multiplying by the reciprocal) to form a product.

Students recognize that any order does not apply to expressions mixing addition and multiplication, leading to the need to follow the order of operations.

Lesson Notes The any order, any grouping property introduced in this lesson combines the commutative and associative properties of both addition and multiplication. The commutative and associative properties are regularly used in sequence to rearrange terms in an expression without necessarily making changes to the terms themselves. Therefore, students utilize the any order, any grouping property as a tool of efficiency for manipulating expressions. The any order, any grouping property is referenced in the Progressions for the Common Core State Standards in Mathematics: Grades 6–8, Expressions and Equations, and is introduced in Grade 6 Module 4, Expressions and Equations.

The definitions presented below, related to variables and expressions, form the foundation of the next few lessons in this topic. Please review these carefully in order to understand the structure of Topic A lessons.

VARIABLE: A variable is a symbol (such as a letter) that represents a number (i.e., it is a placeholder for a number).

A variable is actually quite a simple idea: it is a placeholder—a blank—in an expression or an equation where a number can be inserted. A variable holds a place for a single number throughout all calculations done with the variable—it does not vary. It is the user of the variable who has the ultimate power to change or vary what number is inserted, as he/she desires. The power to vary rests in the will of the student, not in the variable itself.

NUMERICAL EXPRESSION (IN MIDDLE SCHOOL): A numerical expression is a number, or it is any combination of sums, differences, products, or divisions of numbers that evaluates to a number.

Statements such as 3 + or 3 ÷ 0 are not numerical expressions because neither represents a point on the number line.

VALUE OF A NUMERICAL EXPRESSION: The value of a numerical expression is the number found by evaluating the expression.

For example, 13∙ (2 + 4) − 7 is a numerical expression, and its value is −5. Note to teachers: Please do not stress

words over meaning here; it is acceptable to use number computed, computation, calculation, etc. to refer to the value as well.

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7•3 Lesson 1

EXPRESSION (IN MIDDLE SCHOOL): An expression is a numerical expression, or it is the result of replacing some (or all) of the numbers in a numerical expression with variables.

There are two ways to build expressions:

• We can start out with a numerical expression, such as 13∙ (2 + 4) − 7, and replace some of the numbers with

letters to get 13∙ (𝑥𝑥 + 𝑦𝑦) − 𝑧𝑧.

• We can build such expressions from scratch, as in 𝑥𝑥 + 𝑥𝑥(𝑦𝑦 − 𝑧𝑧), and note that if numbers were placed in the expression for the variables 𝑥𝑥, 𝑦𝑦, and 𝑧𝑧, the result would be a numerical expression.

The key is to strongly link expressions back to computations with numbers through building and evaluating them. Building an expression often occurs in the context of a word problem by thinking about examples of numerical expressions first and then replacing some of the numbers with letters in a numerical expression. The act of evaluating an expression means to replace each of the variables with specific numbers to get a numerical expression, and then finding the value of that numerical expression.

The description of expression above is meant to work nicely with how Grade 6 and Grade 7 students learn to manipulate expressions. In these grades, students spend a lot of time building and evaluating expressions for specific numbers substituted into the variables. Building and evaluating helps students see that expressions are really just a slight abstraction of arithmetic in elementary school.

EQUIVALENT EXPRESSIONS (IN MIDDLE SCHOOL): Two expressions are equivalent if both expressions evaluate to the same number for every substitution of numbers into all the letters in both expressions. This description becomes clearer through lots of examples and linking to the associative, commutative, and distributive properties.

AN EXPRESSION IN EXPANDED FORM (IN MIDDLE SCHOOL): An expression that is written as sums (and/or differences) of products whose factors are numbers, variables, or variables raised to whole number powers is said to be in expanded form. A single number, variable, or a single product of numbers and/or variables is also considered to be in expanded form.

AN EXPRESSION IN STANDARD FORM (IN MIDDLE SCHOOL): An expression that is in expanded form where all like terms have been collected is said to be in standard form.

Important: An expression in standard form is the equivalent of what is traditionally referred to as a simplified expression. This curriculum does not utilize the term simplify when writing equivalent expressions, but rather asks students to put an expression in standard form or expand the expression and combine like terms. However, students must know that the term simplify will be seen outside of this curriculum and that the term is directing them to write an expression in standard form.

TERM (DESCRIPTION): Each summand of an expression in expanded form is called a term.

COEFFICIENT OF A TERM: The coefficient of a term is the number found by multiplying all of the number factors in a term together.

Materials

Prepare a classroom set of manila envelopes (non-translucent). Cut and place four triangles and two quadrilaterals in each envelope (provided at the end of this lesson). These envelopes are used in the Opening Exercise of this lesson.

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7•3 Lesson 1

Classwork

Opening Exercise (15 minutes)

This exercise requires students to represent unknown quantities with variable symbols and reason mathematically with those symbols to represent another unknown value.

As students enter the classroom, provide each one with an envelope containing two quadrilaterals and four triangles; instruct students not to open their envelopes. Divide students into teams of two to complete parts (a) and (b).

Opening Exercise

Each envelope contains a number of triangles and a number of quadrilaterals. For this exercise, let 𝒕𝒕 represent the number of triangles, and let 𝒒𝒒 represent the number of quadrilaterals.

a. Write an expression using 𝒕𝒕 and 𝒒𝒒 that represents the total number of sides in your envelope. Explain what the terms in your expression represent.

𝟑𝟑𝒕𝒕+ 𝟒𝟒𝒒𝒒. Triangles have 𝟑𝟑 sides, so there will be 𝟑𝟑 sides for each triangle in the envelope. This is represented by 𝟑𝟑𝒕𝒕. Quadrilaterals have 𝟒𝟒 sides, so there will be 𝟒𝟒 sides for each quadrilateral in the envelope. This is represented by 𝟒𝟒𝒒𝒒. The total number of sides will be the number of triangle sides and the number of quadrilateral sides together.

b. You and your partner have the same number of triangles and quadrilaterals in your envelopes. Write an expression that represents the total number of sides that you and your partner have. If possible, write more than one expression to represent this total.

𝟑𝟑𝒕𝒕+ 𝟒𝟒𝒒𝒒 + 𝟑𝟑𝒕𝒕+ 𝟒𝟒𝒒𝒒; 𝟐𝟐(𝟑𝟑𝒕𝒕+ 𝟒𝟒𝒒𝒒); 𝟔𝟔𝒕𝒕 + 𝟖𝟖𝒒𝒒

Discuss the variations of the expressions in part (b) and whether those variations are equivalent. This discussion helps students understand what it means to combine like terms; some students have added their number of triangles together and number of quadrilaterals together, while others simply doubled their own number of triangles and quadrilaterals since the envelopes contain the same number. This discussion further shows how these different forms of the same expression relate to each other. Students then complete part (c).

c. Each envelope in the class contains the same number of triangles and quadrilaterals. Write an expression that represents the total number of sides in the room.

Answer depends on the number of students in the classroom. For example, if there are 𝟏𝟏𝟐𝟐 students in the classroom, the expression would be 𝟏𝟏𝟐𝟐(𝟑𝟑𝒕𝒕+ 𝟒𝟒𝒒𝒒), or an equivalent expression.

Next, discuss any variations (or possible variations) of the expression in part (c), and discuss whether those variations are equivalent. Are there as many variations in part (c), or did students use multiplication to consolidate the terms in their expressions? If the latter occurred, discuss students’ reasoning.

Scaffolding: To help students understand the given task, discuss a numerical expression, such as

2 × 3 + 6 × 4, as an example where there are two triangles and six quadrilaterals.

MP.2

MP.8

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7•3 Lesson 1

Choose one student to open an envelope and count the numbers of triangles and quadrilaterals. Record the values of 𝑡𝑡 and 𝑞𝑞 as reported by that student for all students to see. Next, students complete parts (d), (e), and (f).

d. Use the given values of 𝒕𝒕 and 𝒒𝒒 and your expression from part (a) to determine the number of sides that should be found in your envelope.

𝟑𝟑𝒕𝒕+ 𝟒𝟒𝒒𝒒 𝟑𝟑(𝟒𝟒) + 𝟒𝟒(𝟐𝟐) 𝟏𝟏𝟐𝟐+ 𝟖𝟖 𝟐𝟐𝟐𝟐

There should be 𝟐𝟐𝟐𝟐 sides contained in my envelope.

e. Use the same values for 𝒕𝒕 and 𝒒𝒒 and your expression from part (b) to determine the number of sides that should be contained in your envelope and your partner’s envelope combined.

Variation 1 Variation 2 Variation 3

𝟐𝟐(𝟑𝟑𝒕𝒕+ 𝟒𝟒𝒒𝒒) 𝟑𝟑𝒕𝒕 + 𝟒𝟒𝒒𝒒 + 𝟑𝟑𝒕𝒕 + 𝟒𝟒𝒒𝒒 𝟔𝟔𝒕𝒕+ 𝟖𝟖𝒒𝒒

𝟐𝟐�𝟑𝟑(𝟒𝟒) + 𝟒𝟒(𝟐𝟐)� 𝟑𝟑(𝟒𝟒) + 𝟒𝟒(𝟐𝟐) + 𝟑𝟑(𝟒𝟒) + 𝟒𝟒(𝟐𝟐) 𝟔𝟔(𝟒𝟒) + 𝟖𝟖(𝟐𝟐) 𝟐𝟐(𝟏𝟏𝟐𝟐+ 𝟖𝟖) 𝟏𝟏𝟐𝟐 + 𝟖𝟖 + 𝟏𝟏𝟐𝟐 + 𝟖𝟖 𝟐𝟐𝟒𝟒+ 𝟏𝟏𝟔𝟔

𝟐𝟐(𝟐𝟐𝟐𝟐) 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 𝟒𝟒𝟐𝟐 𝟒𝟒𝟐𝟐 𝟒𝟒𝟐𝟐

My partner and I have a combined total of 𝟒𝟒𝟐𝟐 sides.

f. Use the same values for 𝒕𝒕 and 𝒒𝒒 and your expression from part (c) to determine the number of sides that should be contained in all of the envelopes combined.

Answer will depend on the seat size of your classroom. Sample responses for a class size of 𝟏𝟏𝟐𝟐:

Variation 1 Variation 2 Variation 3

𝟏𝟏𝟐𝟐(𝟑𝟑𝒕𝒕+ 𝟒𝟒𝒒𝒒) 𝟑𝟑𝒕𝒕 + 𝟒𝟒𝒒𝒒��𝟏𝟏

+ 𝟑𝟑𝒕𝒕 + 𝟒𝟒𝒒𝒒��𝟐𝟐

+ ⋯+ 𝟑𝟑𝒕𝒕+ 𝟒𝟒𝒒𝒒��𝟏𝟏𝟐𝟐

𝟑𝟑𝟔𝟔𝒕𝒕 + 𝟒𝟒𝟖𝟖𝒒𝒒

𝟏𝟏𝟐𝟐�𝟑𝟑(𝟒𝟒) + 𝟒𝟒(𝟐𝟐)� 𝟑𝟑(𝟒𝟒) + 𝟒𝟒(𝟐𝟐)��𝟏𝟏

+ 𝟑𝟑(𝟒𝟒) + 𝟒𝟒(𝟐𝟐)��𝟐𝟐

+ ⋯+ 𝟑𝟑(𝟒𝟒) + 𝟒𝟒(𝟐𝟐)��𝟏𝟏𝟐𝟐

𝟑𝟑𝟔𝟔(𝟒𝟒) + 𝟒𝟒𝟖𝟖(𝟐𝟐)

𝟏𝟏𝟐𝟐(𝟏𝟏𝟐𝟐+ 𝟖𝟖) 𝟏𝟏𝟐𝟐 + 𝟖𝟖��𝟏𝟏

+ 𝟏𝟏𝟐𝟐 + 𝟖𝟖��𝟐𝟐

+ ⋯+ 𝟏𝟏𝟐𝟐+ 𝟖𝟖��𝟏𝟏𝟐𝟐

𝟏𝟏𝟒𝟒𝟒𝟒+ 𝟗𝟗𝟔𝟔

𝟏𝟏𝟐𝟐(𝟐𝟐𝟐𝟐) 𝟐𝟐𝟐𝟐�𝟏𝟏

+ 𝟐𝟐𝟐𝟐�𝟐𝟐

+⋯+ 𝟐𝟐𝟐𝟐�𝟏𝟏𝟐𝟐

𝟐𝟐𝟒𝟒𝟐𝟐

𝟐𝟐𝟒𝟒𝟐𝟐 𝟐𝟐𝟒𝟒𝟐𝟐

For a class size of 𝟏𝟏𝟐𝟐 students, there should be 𝟐𝟐𝟒𝟒𝟐𝟐 sides in all of the envelopes combined.

Have all students open their envelopes and confirm that the number of triangles and quadrilaterals matches the values of 𝑡𝑡 and 𝑞𝑞 recorded after part (c). Then, have students count the number of sides on the triangles and quadrilaterals from their own envelopes and confirm with their answers to part (d). Next, have partners count how many sides they have combined and confirm that number with their answers to part (e). Finally, total the number of sides reported by each student in the classroom and confirm this number with the answer to part (f).

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7•3 Lesson 1

Scaffolding: Refer students to the triangles and quadrilaterals from the Opening Exercise to understand why terms containing the same variable symbol 𝑥𝑥 can be added together into a single term.

g. What do you notice about the various expressions in parts (e) and (f)?

The expressions in part (e) are all equivalent because they evaluate to the same number: 𝟒𝟒𝟐𝟐. The expressions in part (f) are all equivalent because they evaluate to the same number: 𝟐𝟐𝟒𝟒𝟐𝟐. The expressions themselves all involve the expression 𝟑𝟑𝒕𝒕 + 𝟒𝟒𝒒𝒒 in different ways. In part (e), 𝟑𝟑𝒕𝒕+ 𝟑𝟑𝒕𝒕 is equivalent to 𝟔𝟔𝒕𝒕, and 𝟒𝟒𝒒𝒒 + 𝟒𝟒𝒒𝒒 is equivalent to 𝟖𝟖𝒒𝒒. There appear to be several relationships among the representations involving the commutative, associative, and distributive properties.

When finished, have students return their triangles and quadrilaterals to their envelopes for use by other classes.

Example 1 (10 minutes): Any Order, Any Grouping Property with Addition

This example examines how and why we combine numbers and other like terms in expressions. An expression that is written as sums (and/or differences) of products whose factors are numbers, variables, or variables raised to whole number powers is said to be in expanded form. A single number, variable, or a single product of numbers and/or variables is also considered to be in expanded form. Examples of expressions in expanded form include 324, 3𝑥𝑥, 5𝑥𝑥 + 3 − 40, 𝑥𝑥 + 2𝑥𝑥 + 3𝑥𝑥, etc.

Each summand of an expression in expanded form is called a term, and the number found by multiplying just the numbers in a term together is called the coefficient of the term. After defining the word term, students can be shown what it means to combine like terms using the distributive property. Students saw in the Opening Exercise that terms sharing exactly the same letter could be combined by adding (or subtracting) the coefficients of the terms:

3𝑡𝑡 + 3𝑡𝑡 = (3 + 3)��coefficients

∙ 𝑡𝑡 = 6𝑡𝑡, and 4𝑞𝑞 + 4𝑞𝑞 = (4 + 4)��coefficients

∙ 𝑞𝑞 = 8𝑞𝑞.

An expression in expanded form with all its like terms collected is said to be in standard form.

Example 1: Any Order, Any Grouping Property with Addition

a. Rewrite 𝟓𝟓𝒙𝒙 + 𝟑𝟑𝒙𝒙 and 𝟓𝟓𝒙𝒙 − 𝟑𝟑𝒙𝒙 by combining like terms.

Write the original expressions and expand each term using addition. What are the new expressions equivalent to?

𝟓𝟓𝒙𝒙 + 𝟑𝟑𝒙𝒙 = 𝒙𝒙 + 𝒙𝒙 + 𝒙𝒙 + 𝒙𝒙 + 𝒙𝒙��𝟓𝟓𝒙𝒙

+ 𝒙𝒙 + 𝒙𝒙 + 𝒙𝒙��𝟑𝟑𝒙𝒙

��𝟖𝟖𝒙𝒙

= 𝟖𝟖𝒙𝒙

𝟓𝟓𝒙𝒙 − 𝟑𝟑𝒙𝒙 = 𝒙𝒙 + 𝒙𝒙 + 𝒙𝒙 + 𝒙𝒙 + 𝒙𝒙��𝟑𝟑𝒙𝒙

��𝟓𝟓𝒙𝒙

= 𝟐𝟐𝒙𝒙

Because both terms have the common factor of 𝑥𝑥, we can use the distributive property to create an equivalent expression.

𝟓𝟓𝒙𝒙 + 𝟑𝟑𝒙𝒙 = (𝟓𝟓+ 𝟑𝟑)𝒙𝒙 = 𝟖𝟖𝒙𝒙 𝟓𝟓𝒙𝒙 − 𝟑𝟑𝒙𝒙 = (𝟓𝟓 − 𝟑𝟑)𝒙𝒙 = 𝟐𝟐𝒙𝒙

Ask students to try to find an example (a value for 𝑥𝑥) where 5𝑥𝑥 + 3𝑥𝑥 ≠ 8𝑥𝑥 or where 5𝑥𝑥 − 3𝑥𝑥 ≠ 2𝑥𝑥. Encourage them to use a variety of positive and negative rational numbers. Their failure to find a counterexample helps students realize what equivalence means.

Scaffolding: Note to the teacher: The distributive property was covered in Grade 6 (6.EE.A.3) and is reviewed here in preparation for further use in this module starting with Lesson 3.

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7•3 Lesson 1

In Example 1, part (b), students see that the commutative and associative properties of addition are regularly used in consecutive steps to reorder and regroup like terms so that they can be combined. Because the use of these properties does not change the value of an expression or any of the terms within the expression, the commutative and associative properties of addition can be used simultaneously. The simultaneous use of these properties is referred to as the any order, any grouping property.

b. Find the sum of 𝟐𝟐𝒙𝒙 + 𝟏𝟏 and 𝟓𝟓𝒙𝒙.

(𝟐𝟐𝒙𝒙 + 𝟏𝟏) + 𝟓𝟓𝒙𝒙 Original expression

𝟐𝟐𝒙𝒙 + (𝟏𝟏+ 𝟓𝟓𝒙𝒙) Associative property of addition

𝟐𝟐𝒙𝒙 + (𝟓𝟓𝒙𝒙+ 𝟏𝟏) Commutative property of addition

(𝟐𝟐𝒙𝒙 + 𝟓𝟓𝒙𝒙) + 𝟏𝟏 Associative property of addition

(𝟐𝟐+ 𝟓𝟓)𝒙𝒙 + 𝟏𝟏 Combined like terms (the distributive property)

𝟕𝟕𝒙𝒙 + 𝟏𝟏 Equivalent expression to the given problem

Why did we use the associative and commutative properties of addition?

We reordered the terms in the expression to group together like terms so that they could be combined.

Did the use of these properties change the value of the expression? How do you know?

The properties did not change the value of the expression because each equivalent expression includes the same terms as the original expression, just in a different order and grouping.

If a sequence of terms is being added, the any order, any grouping property allows us to add those terms in any order by grouping them together in any way.

How can we confirm that the expressions (2𝑥𝑥 + 1) + 5𝑥𝑥 and 7𝑥𝑥 + 1 are equivalent expressions?

When a number is substituted for the 𝑥𝑥 in both expressions, they both should yield equal results.

The teacher and student should choose a number, such as 3, to substitute for the value of 𝑥𝑥 and together check to see if both expressions evaluate to the same result.

Given Expression Equivalent Expression?

(𝟐𝟐𝒙𝒙+ 𝟏𝟏) + 𝟓𝟓𝒙𝒙(𝟐𝟐 ∙ 𝟑𝟑+ 𝟏𝟏) + 𝟓𝟓 ∙ 𝟑𝟑

(𝟔𝟔+ 𝟏𝟏) + 𝟏𝟏𝟓𝟓(𝟕𝟕) + 𝟏𝟏𝟓𝟓

𝟐𝟐𝟐𝟐

𝟕𝟕𝒙𝒙 + 𝟏𝟏𝟕𝟕 ∙ 𝟑𝟑 + 𝟏𝟏𝟐𝟐𝟏𝟏 + 𝟏𝟏𝟐𝟐𝟐𝟐

The expressions both evaluate to 22; however, this is only one possible value of 𝑥𝑥. Challenge students to find a value for 𝑥𝑥 for which the expressions do not yield the same number. Students find that the expressions evaluate to equal results no matter what value is chosen for 𝑥𝑥.

What prevents us from using any order, any grouping in part (c), and what can we do about it?

The second expression, (5𝑎𝑎 − 3), involves subtraction, which is not commutative or associative; however, subtracting a number 𝑥𝑥 can be written as adding the opposite of that number. So, by changing subtraction to addition, we can use any order and any grouping.

With a firm understanding of the commutative and associative properties of addition, students further understand that these steps can be combined.

Scaffolding: Teacher may also want to show the expression as 𝑥𝑥 + 𝑥𝑥 + 1��

2𝑥𝑥+1+ 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥��

5𝑥𝑥

in the same manner as in part (a).

MP.7

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7•3 Lesson 1

c. Find the sum of −𝟑𝟑𝒂𝒂 + 𝟐𝟐 and 𝟓𝟓𝒂𝒂 − 𝟑𝟑.

(−𝟑𝟑𝒂𝒂+ 𝟐𝟐) + (𝟓𝟓𝒂𝒂 − 𝟑𝟑) Original expression −𝟑𝟑𝒂𝒂 + 𝟐𝟐 + 𝟓𝟓𝒂𝒂 + (−𝟑𝟑) Add the opposite (additive inverse) −𝟑𝟑𝒂𝒂 + 𝟓𝟓𝒂𝒂 + 𝟐𝟐 + (−𝟑𝟑) Any order, any grouping 𝟐𝟐𝒂𝒂 + (−𝟏𝟏) Combined like terms (Stress to students that the expression is not yet simplified.) 𝟐𝟐𝒂𝒂 − 𝟏𝟏 Adding the inverse is subtracting.

What was the only difference between this problem and those involving all addition?

We first had to rewrite subtraction as addition; then, this problem was just like the others.

Example 2 (3 minutes): Any Order, Any Grouping with Multiplication

Students relate a product to its expanded form and understand that the same result can be obtained using any order, any grouping since multiplication is also associative and commutative.

Example 2: Any Order, Any Grouping with Multiplication

Find the product of 𝟐𝟐𝒙𝒙 and 𝟑𝟑.

𝟐𝟐𝒙𝒙 ∙ 𝟑𝟑

𝟐𝟐 ∙ (𝒙𝒙 ∙ 𝟑𝟑) Associative property of multiplication (any grouping)

𝟐𝟐 ∙ (𝟑𝟑 ∙ 𝒙𝒙) Commutative property of multiplication (any order)

𝟔𝟔𝒙𝒙 Multiplication

Why did we use the associative and commutative properties of multiplication? We reordered the factors to group together the numbers so that they could be multiplied.

Did the use of these properties change the value of the expression? How do you know?

The properties did not change the value of the expression because each equivalent expression includes the same factors as the original expression, just in a different order or grouping.

If a product of factors is being multiplied, the any order, any grouping property allows us to multiply those factors in any order by grouping them together in any way.

Example 3 (9 minutes): Any Order, Any Grouping in Expressions with Addition and Multiplication

Students use any order, any grouping to rewrite products with a single coefficient first as terms only, then as terms within a sum, noticing that any order, any grouping cannot be used to mix multiplication with addition.

Example 3: Any Order, Any Grouping in Expressions with Addition and Multiplication

Use any order, any grouping to write equivalent expressions.

a. 𝟑𝟑(𝟐𝟐𝒙𝒙)

(𝟑𝟑 ∙ 𝟐𝟐)𝒙𝒙

𝟔𝟔𝒙𝒙

With a firm understanding of the commutative and associative properties of multiplication, students further understand that these steps can be combined. MP.7

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7•3 Lesson 1

Ask students to try to find an example (a value for 𝑥𝑥) where 3(2𝑥𝑥) ≠ 6𝑥𝑥. Encourage them to use a variety of positive and negative rational numbers because in order for the expressions to be equivalent, the expressions must evaluate to equal numbers for every substitution of numbers into all the letters in both expressions. Again, the point is to help students recognize that they cannot find a value—that the two expressions are equivalent. Encourage students to follow the order of operations for the expression 3(2𝑥𝑥): multiply by 2 first, then by 3.

b. 𝟒𝟒𝟒𝟒(𝟓𝟓)

(𝟒𝟒 ∙ 𝟓𝟓)𝟒𝟒

𝟐𝟐𝟐𝟐𝟒𝟒

c. 𝟒𝟒 ∙ 𝟐𝟐 ∙ 𝒛𝒛

(𝟒𝟒 ∙ 𝟐𝟐)𝒛𝒛

𝟖𝟖𝒛𝒛

d. 𝟑𝟑(𝟐𝟐𝒙𝒙) + 𝟒𝟒𝟒𝟒(𝟓𝟓)

𝟑𝟑(𝟐𝟐𝒙𝒙) + 𝟒𝟒𝟒𝟒(𝟓𝟓) = 𝟐𝟐𝒙𝒙 + 𝟐𝟐𝒙𝒙 + 𝟐𝟐𝒙𝒙��𝟔𝟔𝒙𝒙

+ 𝟒𝟒𝟒𝟒+ 𝟒𝟒𝟒𝟒 + 𝟒𝟒𝟒𝟒 + 𝟒𝟒𝟒𝟒 + 𝟒𝟒𝟒𝟒��𝟐𝟐𝟐𝟐𝟒𝟒

(𝟑𝟑 ∙ 𝟐𝟐)𝒙𝒙 + (𝟒𝟒 ∙ 𝟓𝟓)𝟒𝟒

𝟔𝟔𝒙𝒙 + 𝟐𝟐𝟐𝟐𝟒𝟒

e. 𝟑𝟑(𝟐𝟐𝒙𝒙) + 𝟒𝟒𝟒𝟒(𝟓𝟓) + 𝟒𝟒 ∙ 𝟐𝟐 ∙ 𝒛𝒛

𝟑𝟑(𝟐𝟐𝒙𝒙) + 𝟒𝟒𝟒𝟒(𝟓𝟓) + 𝟒𝟒 ∙ 𝟐𝟐 ∙ 𝒛𝒛 = 𝟐𝟐𝒙𝒙 + 𝟐𝟐𝒙𝒙 + 𝟐𝟐𝒙𝒙��𝟔𝟔𝒙𝒙

+ 𝟒𝟒𝟒𝟒+ 𝟒𝟒𝟒𝟒+ 𝟒𝟒𝟒𝟒 + 𝟒𝟒𝟒𝟒 + 𝟒𝟒𝟒𝟒��𝟐𝟐𝟐𝟐𝟒𝟒

+ 𝒛𝒛 + 𝒛𝒛 + 𝒛𝒛 + 𝒛𝒛 + 𝒛𝒛 + 𝒛𝒛 + 𝒛𝒛 + 𝒛𝒛��𝟖𝟖𝒛𝒛

(𝟑𝟑 ∙ 𝟐𝟐)𝒙𝒙 + (𝟒𝟒 ∙ 𝟓𝟓)𝟒𝟒+ (𝟒𝟒 ∙ 𝟐𝟐)𝒛𝒛

𝟔𝟔𝒙𝒙 + 𝟐𝟐𝟐𝟐𝟒𝟒+ 𝟖𝟖𝒛𝒛

f. Alexander says that 𝟑𝟑𝒙𝒙 + 𝟒𝟒𝟒𝟒 is equivalent to (𝟑𝟑)(𝟒𝟒) + 𝒙𝒙𝟒𝟒 because of any order, any grouping. Is he correct? Why or why not?

Encourage students to substitute a variety of positive and negative rational numbers for 𝑥𝑥 and 𝑦𝑦 because in order for the expressions to be equivalent, the expressions must evaluate to equal numbers for every substitution of numbers into all the letters in both expressions.

Alexander is incorrect; the expressions are not equivalent because if we, for example, let 𝒙𝒙 = −𝟐𝟐 and let 𝟒𝟒 = −𝟑𝟑, then we get the following:

𝟑𝟑𝒙𝒙 + 𝟒𝟒𝟒𝟒 (𝟑𝟑)(𝟒𝟒) + 𝒙𝒙𝟒𝟒

𝟑𝟑(−𝟐𝟐) + 𝟒𝟒(−𝟑𝟑) 𝟏𝟏𝟐𝟐 + (−𝟐𝟐)(−𝟑𝟑)

−𝟔𝟔+ (−𝟏𝟏𝟐𝟐) 𝟏𝟏𝟐𝟐 + 𝟔𝟔

−𝟏𝟏𝟖𝟖 𝟏𝟏𝟖𝟖

−𝟏𝟏𝟖𝟖 ≠ 𝟏𝟏𝟖𝟖, so the expressions cannot be equivalent.

What can be concluded as a result of part (f)?

Any order, any grouping cannot be used to mix multiplication with addition. Numbers and letters that are factors within a given term must remain factors within that term.

MP.3

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7•3 Lesson 1

Closing (3 minutes)

We found that we can use any order, any grouping of terms in a sum, or of factors in a product. Why?

Addition and multiplication are both associative and commutative, and these properties only change the order and grouping of terms in a sum or factors in a product without affecting the value of the expression.

Can we use any order, any grouping when subtracting expressions? Explain.

We can use any order, any grouping after rewriting subtraction as the sum of a number and the additive inverse of that number so that the expression becomes a sum.

Why can’t we use any order, any grouping in addition and multiplication at the same time? Multiplication must be completed before addition. If you mix the operations, you change the value of

the expression.

Relevant Vocabulary

VARIABLE (DESCRIPTION): A variable is a symbol (such as a letter) that represents a number (i.e., it is a placeholder for a number).

NUMERICAL EXPRESSION (DESCRIPTION): A numerical expression is a number, or it is any combination of sums, differences, products, or divisions of numbers that evaluates to a number.

VALUE OF A NUMERICAL EXPRESSION: The value of a numerical expression is the number found by evaluating the expression.

EXPRESSION (DESCRIPTION): An expression is a numerical expression, or it is the result of replacing some (or all) of the numbers in a numerical expression with variables.

EQUIVALENT EXPRESSIONS: Two expressions are equivalent if both expressions evaluate to the same number for every substitution of numbers into all the letters in both expressions.

AN EXPRESSION IN EXPANDED FORM: An expression that is written as sums (and/or differences) of products whose factors are numbers, variables, or variables raised to whole number powers is said to be in expanded form. A single number, variable, or a single product of numbers and/or variables is also considered to be in expanded form. Examples of expressions in expanded form include: 𝟑𝟑𝟐𝟐𝟒𝟒, 𝟑𝟑𝒙𝒙, 𝟓𝟓𝒙𝒙+ 𝟑𝟑 − 𝟒𝟒𝟐𝟐, and 𝒙𝒙 + 𝟐𝟐𝒙𝒙 + 𝟑𝟑𝒙𝒙.

TERM (DESCRIPTION): Each summand of an expression in expanded form is called a term. For example, the expression 𝟐𝟐𝒙𝒙 + 𝟑𝟑𝒙𝒙 + 𝟓𝟓 consists of three terms: 𝟐𝟐𝒙𝒙, 𝟑𝟑𝒙𝒙, and 𝟓𝟓.

COEFFICIENT OF THE TERM (DESCRIPTION): The number found by multiplying just the numbers in a term together is the coefficient of the term. For example, given the product 𝟐𝟐 ∙ 𝒙𝒙 ∙ 𝟒𝟒, its equivalent term is 𝟖𝟖𝒙𝒙. The number 𝟖𝟖 is called the coefficient of the term 𝟖𝟖𝒙𝒙.

AN EXPRESSION IN STANDARD FORM: An expression in expanded form with all its like terms collected is said to be in standard form. For example, 𝟐𝟐𝒙𝒙 + 𝟑𝟑𝒙𝒙 + 𝟓𝟓 is an expression written in expanded form; however, to be written in standard form, the like terms 𝟐𝟐𝒙𝒙 and 𝟑𝟑𝒙𝒙 must be combined. The equivalent expression 𝟓𝟓𝒙𝒙 + 𝟓𝟓 is written in standard form.

Exit Ticket (5 minutes)

Lesson Summary

Terms that contain exactly the same variable symbol can be combined by addition or subtraction because the variable represents the same number. Any order, any grouping can be used where terms are added (or subtracted) in order to group together like terms. Changing the orders of the terms in a sum does not affect the value of the expression for given values of the variable(s).

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7•3 Lesson 1

Name ___________________________________________________ Date____________________


Exit Ticket 1. Write an equivalent expression to 2𝑥𝑥 + 3 + 5𝑥𝑥 + 6 by combining like terms.

2. Find the sum of (8𝑎𝑎 + 2𝑏𝑏 − 4) and (3𝑏𝑏 − 5).

3. Write the expression in standard form: 4(2𝑎𝑎) + 7(−4𝑏𝑏) + (3 ∙ 𝑐𝑐 ∙ 5).

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7•3 Lesson 1

Exit Ticket Sample Solutions

1. Write an equivalent expression to 𝟐𝟐𝒙𝒙 + 𝟑𝟑 + 𝟓𝟓𝒙𝒙 + 𝟔𝟔 by combining like terms.

𝟐𝟐𝒙𝒙 + 𝟑𝟑+ 𝟓𝟓𝒙𝒙 + 𝟔𝟔

𝟐𝟐𝒙𝒙 + 𝟓𝟓𝒙𝒙 + 𝟑𝟑+ 𝟔𝟔

𝟕𝟕𝒙𝒙 + 𝟗𝟗

2. Find the sum of (𝟖𝟖𝒂𝒂 + 𝟐𝟐𝒃𝒃 − 𝟒𝟒) and (𝟑𝟑𝒃𝒃 − 𝟓𝟓).

(𝟖𝟖𝒂𝒂 + 𝟐𝟐𝒃𝒃 − 𝟒𝟒) + (𝟑𝟑𝒃𝒃 − 𝟓𝟓)

𝟖𝟖𝒂𝒂 + 𝟐𝟐𝒃𝒃 + (−𝟒𝟒) + 𝟑𝟑𝒃𝒃 + (−𝟓𝟓)

𝟖𝟖𝒂𝒂 + 𝟐𝟐𝒃𝒃 + 𝟑𝟑𝒃𝒃 + (−𝟒𝟒) + (−𝟓𝟓)

𝟖𝟖𝒂𝒂 + (𝟓𝟓𝒃𝒃) + (−𝟗𝟗)

𝟖𝟖𝒂𝒂 + 𝟓𝟓𝒃𝒃 − 𝟗𝟗

3. Write the expression in standard form: 𝟒𝟒(𝟐𝟐𝒂𝒂) + 𝟕𝟕(−𝟒𝟒𝒃𝒃) + (𝟑𝟑 ∙ 𝒄𝒄 ∙ 𝟓𝟓).

(𝟒𝟒 ∙ 𝟐𝟐)𝒂𝒂+ �𝟕𝟕 ∙ (−𝟒𝟒)�𝒃𝒃+ (𝟑𝟑 ∙ 𝟓𝟓)𝒄𝒄

𝟖𝟖𝒂𝒂 + (−𝟐𝟐𝟖𝟖)𝒃𝒃+ 𝟏𝟏𝟓𝟓𝒄𝒄

𝟖𝟖𝒂𝒂 − 𝟐𝟐𝟖𝟖𝒃𝒃 + 𝟏𝟏𝟓𝟓𝒄𝒄

Problem Set Sample Solutions

For Problems 1–9, write equivalent expressions by combining like terms. Verify the equivalence of your expression and the given expression by evaluating each for the given values: 𝒂𝒂 = 𝟐𝟐, 𝒃𝒃 = 𝟓𝟓, and 𝒄𝒄 = −𝟑𝟑.

1. 𝟑𝟑𝒂𝒂 + 𝟓𝟓𝒂𝒂 2. 𝟖𝟖𝒃𝒃 − 𝟒𝟒𝒃𝒃 3. 𝟓𝟓𝒄𝒄 + 𝟒𝟒𝒄𝒄 + 𝒄𝒄

𝟖𝟖𝒂𝒂 𝟖𝟖(𝟐𝟐) 𝟏𝟏𝟔𝟔

𝟑𝟑(𝟐𝟐) + 𝟓𝟓(𝟐𝟐) 𝟔𝟔 + 𝟏𝟏𝟐𝟐 𝟏𝟏𝟔𝟔

𝟒𝟒𝒃𝒃 𝟒𝟒(𝟓𝟓) 𝟐𝟐𝟐𝟐 𝟖𝟖(𝟓𝟓) − 𝟒𝟒(𝟓𝟓) 𝟒𝟒𝟐𝟐 − 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐

𝟏𝟏𝟐𝟐𝒄𝒄 𝟏𝟏𝟐𝟐(−𝟑𝟑) −𝟑𝟑𝟐𝟐 𝟓𝟓(−𝟑𝟑) + 𝟒𝟒(−𝟑𝟑) + (−𝟑𝟑) −𝟏𝟏𝟓𝟓+ (−𝟏𝟏𝟐𝟐) + (−𝟑𝟑) −𝟐𝟐𝟕𝟕+ (−𝟑𝟑) −𝟑𝟑𝟐𝟐

4. 𝟑𝟑𝒂𝒂 + 𝟔𝟔+ 𝟓𝟓𝒂𝒂 5. 𝟖𝟖𝒃𝒃 + 𝟖𝟖 − 𝟒𝟒𝒃𝒃 6. 𝟓𝟓𝒄𝒄 − 𝟒𝟒𝒄𝒄 + 𝒄𝒄

𝟖𝟖𝒂𝒂 + 𝟔𝟔 𝟖𝟖(𝟐𝟐) + 𝟔𝟔 𝟏𝟏𝟔𝟔 + 𝟔𝟔 𝟐𝟐𝟐𝟐 𝟑𝟑(𝟐𝟐) + 𝟔𝟔+ 𝟓𝟓(𝟐𝟐) 𝟔𝟔 + 𝟔𝟔 + 𝟏𝟏𝟐𝟐 𝟏𝟏𝟐𝟐 + 𝟏𝟏𝟐𝟐 𝟐𝟐𝟐𝟐

𝟒𝟒𝒃𝒃 + 𝟖𝟖 𝟒𝟒(𝟓𝟓) + 𝟖𝟖 𝟐𝟐𝟐𝟐 + 𝟖𝟖 𝟐𝟐𝟖𝟖 𝟖𝟖(𝟓𝟓) + 𝟖𝟖 − 𝟒𝟒(𝟓𝟓) 𝟒𝟒𝟐𝟐 + 𝟖𝟖 − 𝟐𝟐𝟐𝟐 𝟒𝟒𝟖𝟖 − 𝟐𝟐𝟐𝟐 𝟐𝟐𝟖𝟖

𝟐𝟐𝒄𝒄 𝟐𝟐(−𝟑𝟑) −𝟔𝟔 𝟓𝟓(−𝟑𝟑) − 𝟒𝟒(−𝟑𝟑) + (−𝟑𝟑) −𝟏𝟏𝟓𝟓+ �−𝟒𝟒(−𝟑𝟑)�+ (−𝟑𝟑) −𝟏𝟏𝟓𝟓+ (𝟏𝟏𝟐𝟐) + (−𝟑𝟑) −𝟑𝟑+ (−𝟑𝟑) −𝟔𝟔

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7•3 Lesson 1

7. 𝟑𝟑𝒂𝒂 + 𝟔𝟔+ 𝟓𝟓𝒂𝒂 − 𝟐𝟐 8. 𝟖𝟖𝒃𝒃 + 𝟖𝟖 − 𝟒𝟒𝒃𝒃 − 𝟑𝟑 9. 𝟓𝟓𝒄𝒄 − 𝟒𝟒𝒄𝒄 + 𝒄𝒄 − 𝟑𝟑𝒄𝒄

𝟖𝟖𝒂𝒂 + 𝟒𝟒 𝟖𝟖(𝟐𝟐) + 𝟒𝟒 𝟏𝟏𝟔𝟔 + 𝟒𝟒 𝟐𝟐𝟐𝟐 𝟑𝟑(𝟐𝟐) + 𝟔𝟔+ 𝟓𝟓(𝟐𝟐) − 𝟐𝟐 𝟔𝟔 + 𝟔𝟔 + 𝟏𝟏𝟐𝟐+ (−𝟐𝟐) 𝟏𝟏𝟐𝟐 + 𝟏𝟏𝟐𝟐 + (−𝟐𝟐) 𝟐𝟐𝟐𝟐 + (−𝟐𝟐) 𝟐𝟐𝟐𝟐

𝟒𝟒𝒃𝒃 + 𝟓𝟓 𝟒𝟒(𝟓𝟓) + 𝟓𝟓 𝟐𝟐𝟐𝟐 + 𝟓𝟓 𝟐𝟐𝟓𝟓 𝟖𝟖(𝟓𝟓) + 𝟖𝟖 − 𝟒𝟒(𝟓𝟓) − 𝟑𝟑 𝟒𝟒𝟐𝟐 + 𝟖𝟖 + �−𝟒𝟒(𝟓𝟓)�+ (−𝟑𝟑) 𝟒𝟒𝟐𝟐 + 𝟖𝟖 + (−𝟐𝟐𝟐𝟐) + (−𝟑𝟑) 𝟒𝟒𝟖𝟖 + (−𝟐𝟐𝟐𝟐) + (−𝟑𝟑) 𝟐𝟐𝟖𝟖 + (−𝟑𝟑) 𝟐𝟐𝟓𝟓

−𝟏𝟏𝒄𝒄 −𝟏𝟏(−𝟑𝟑) 𝟑𝟑 𝟓𝟓(−𝟑𝟑) − 𝟒𝟒(−𝟑𝟑) + (−𝟑𝟑)− 𝟑𝟑(−𝟑𝟑) −𝟏𝟏𝟓𝟓+ �−𝟒𝟒(−𝟑𝟑)�+ (−𝟑𝟑) + �−𝟑𝟑(−𝟑𝟑)� −𝟏𝟏𝟓𝟓+ (𝟏𝟏𝟐𝟐) + (−𝟑𝟑) + (𝟗𝟗) −𝟑𝟑+ (−𝟑𝟑) + 𝟗𝟗 −𝟔𝟔+ 𝟗𝟗 𝟑𝟑

Use any order, any grouping to write equivalent expressions by combining like terms. Then, verify the equivalence of your expression to the given expression by evaluating for the value(s) given in each problem.

Problem Your Expression Given Expression

10. 𝟑𝟑(𝟔𝟔𝒂𝒂); for 𝒂𝒂 = 𝟑𝟑

𝟏𝟏𝟖𝟖𝒂𝒂

𝟏𝟏𝟖𝟖𝒂𝒂 𝟏𝟏𝟖𝟖(𝟑𝟑) 𝟓𝟓𝟒𝟒

𝟑𝟑�𝟔𝟔(𝟑𝟑)� 𝟑𝟑(𝟏𝟏𝟖𝟖) 𝟓𝟓𝟒𝟒

11. 𝟓𝟓𝟓𝟓(𝟒𝟒); for 𝟓𝟓 = −𝟐𝟐

𝟐𝟐𝟐𝟐𝟓𝟓

𝟐𝟐𝟐𝟐𝟓𝟓 𝟐𝟐𝟐𝟐(−𝟐𝟐) −𝟒𝟒𝟐𝟐

𝟓𝟓(−𝟐𝟐)(𝟒𝟒) −𝟏𝟏𝟐𝟐(𝟒𝟒) −𝟒𝟒𝟐𝟐

12. (𝟓𝟓𝟓𝟓)(−𝟐𝟐); for 𝟓𝟓 = −𝟑𝟑

−𝟏𝟏𝟐𝟐𝟓𝟓

−𝟏𝟏𝟐𝟐𝟓𝟓 −𝟏𝟏𝟐𝟐(−𝟑𝟑) 𝟑𝟑𝟐𝟐

�𝟓𝟓(−𝟑𝟑)�(−𝟐𝟐) (−𝟏𝟏𝟓𝟓)(−𝟐𝟐) 𝟑𝟑𝟐𝟐

13. 𝟑𝟑𝒃𝒃(𝟖𝟖) + (−𝟐𝟐)(𝟕𝟕𝒄𝒄); for 𝒃𝒃 = 𝟐𝟐, 𝒄𝒄 = 𝟑𝟑

𝟐𝟐𝟒𝟒𝒃𝒃 − 𝟏𝟏𝟒𝟒𝒄𝒄

𝟐𝟐𝟒𝟒𝒃𝒃 − 𝟏𝟏𝟒𝟒𝒄𝒄 𝟐𝟐𝟒𝟒(𝟐𝟐) − 𝟏𝟏𝟒𝟒(𝟑𝟑) 𝟒𝟒𝟖𝟖 − 𝟒𝟒𝟐𝟐 𝟔𝟔

𝟑𝟑(𝟐𝟐)(𝟖𝟖) + (−𝟐𝟐)(𝟕𝟕(𝟑𝟑)) 𝟔𝟔(𝟖𝟖) + (−𝟐𝟐)(𝟐𝟐𝟏𝟏) 𝟒𝟒𝟖𝟖 + (−𝟒𝟒𝟐𝟐) 𝟔𝟔

14. −𝟒𝟒(𝟑𝟑𝟑𝟑) + 𝟐𝟐(−𝒕𝒕); for 𝟑𝟑 = 𝟏𝟏𝟐𝟐, 𝒕𝒕 = −𝟑𝟑

−𝟏𝟏𝟐𝟐𝟑𝟑 − 𝟐𝟐𝒕𝒕

−𝟏𝟏𝟐𝟐𝟑𝟑 − 𝟐𝟐𝒕𝒕 −𝟏𝟏𝟐𝟐�𝟏𝟏𝟐𝟐� − 𝟐𝟐(−𝟑𝟑)

−𝟔𝟔+ �−𝟐𝟐(−𝟑𝟑)� −𝟔𝟔+ (𝟔𝟔) 𝟐𝟐

−𝟒𝟒�𝟑𝟑�𝟏𝟏𝟐𝟐��+ 𝟐𝟐�−(−𝟑𝟑)�

−𝟒𝟒�𝟑𝟑𝟐𝟐�+ 𝟐𝟐(𝟑𝟑)

−𝟐𝟐(𝟑𝟑) + 𝟐𝟐(𝟑𝟑) −𝟔𝟔+ 𝟔𝟔

𝟐𝟐

15. 𝟗𝟗(𝟒𝟒𝟒𝟒) − 𝟐𝟐(𝟑𝟑𝒒𝒒) + 𝟒𝟒; for 𝟒𝟒 = −𝟏𝟏, 𝒒𝒒 = 𝟒𝟒

𝟑𝟑𝟕𝟕𝟒𝟒 − 𝟔𝟔𝒒𝒒

𝟑𝟑𝟕𝟕𝟒𝟒 − 𝟔𝟔𝒒𝒒 𝟑𝟑𝟕𝟕(−𝟏𝟏) − 𝟔𝟔(𝟒𝟒) −𝟑𝟑𝟕𝟕+ �−𝟔𝟔(𝟒𝟒)� −𝟑𝟑𝟕𝟕+ (−𝟐𝟐𝟒𝟒) −𝟔𝟔𝟏𝟏

𝟗𝟗�𝟒𝟒(−𝟏𝟏)� − 𝟐𝟐�𝟑𝟑(𝟒𝟒)�+ (−𝟏𝟏) 𝟗𝟗(−𝟒𝟒) + �−𝟐𝟐(𝟏𝟏𝟐𝟐)�+ (−𝟏𝟏) −𝟑𝟑𝟔𝟔+ (−𝟐𝟐𝟒𝟒) + (−𝟏𝟏) −𝟔𝟔𝟐𝟐+ (−𝟏𝟏) −𝟔𝟔𝟏𝟏

16. 𝟕𝟕(𝟒𝟒𝟒𝟒) + 𝟑𝟑(𝟓𝟓𝟓𝟓) + 𝟐𝟐(−𝟑𝟑𝟒𝟒); for 𝟒𝟒 = 𝟏𝟏𝟐𝟐, 𝟓𝟓 = 𝟏𝟏

𝟑𝟑

𝟐𝟐𝟖𝟖𝟒𝟒+ 𝟏𝟏𝟓𝟓𝟓𝟓 + (−𝟔𝟔𝟒𝟒) 𝟐𝟐𝟐𝟐𝟒𝟒+ 𝟏𝟏𝟓𝟓𝟓𝟓

𝟐𝟐𝟐𝟐𝟒𝟒 + 𝟏𝟏𝟓𝟓𝟓𝟓

𝟐𝟐𝟐𝟐�𝟏𝟏𝟐𝟐�+ 𝟏𝟏𝟓𝟓�𝟏𝟏𝟑𝟑� 𝟏𝟏𝟏𝟏 + 𝟓𝟓 𝟏𝟏𝟔𝟔

𝟕𝟕�𝟒𝟒 �𝟏𝟏𝟐𝟐��+ 𝟑𝟑�𝟓𝟓 �𝟏𝟏𝟑𝟑��+ 𝟐𝟐�−𝟑𝟑�𝟏𝟏𝟐𝟐��

𝟕𝟕(𝟐𝟐) + 𝟑𝟑�𝟓𝟓𝟑𝟑�+ 𝟐𝟐�−𝟑𝟑𝟐𝟐�

𝟏𝟏𝟒𝟒 + 𝟓𝟓 + (−𝟑𝟑) 𝟏𝟏𝟗𝟗 + (−𝟑𝟑)

𝟏𝟏𝟔𝟔

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7•3 Lesson 1

The problems below are follow-up questions to Example 1, part (b) from Classwork: Find the sum of 𝟐𝟐𝒙𝒙 + 𝟏𝟏 and 𝟓𝟓𝒙𝒙.

17. Jack got the expression 𝟕𝟕𝒙𝒙 + 𝟏𝟏 and then wrote his answer as 𝟏𝟏+ 𝟕𝟕𝒙𝒙. Is his answer an equivalent expression? How do you know?

Yes; Jack correctly applied any order (the commutative property), changing the order of addition.

18. Jill also got the expression 𝟕𝟕𝒙𝒙 + 𝟏𝟏 and then wrote her answer as 𝟏𝟏𝒙𝒙 + 𝟕𝟕. Is her expression an equivalent expression? How do you know?

No, any order (the commutative property) does not apply to mixing addition and multiplication; therefore, the 𝟕𝟕𝒙𝒙 must remain intact as a term.

𝟏𝟏(𝟒𝟒) + 𝟕𝟕 = 𝟏𝟏𝟏𝟏 and 𝟕𝟕(𝟒𝟒) + 𝟏𝟏 = 𝟐𝟐𝟗𝟗; the expressions do not evaluate to the same value for 𝒙𝒙 = 𝟒𝟒.

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7•3 Lesson 1

Materials for Opening Exercise Copy each page and cut out the triangles and quadrilaterals for use in the Opening Exercise.

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7•3 Lesson 1

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7•3 Lesson 2


Student Outcomes

Students generate equivalent expressions using the fact that addition and multiplication can be done in any order (commutative property) and any grouping (associative property).

Students recognize how any order, any grouping can be applied in a subtraction problem by using additive inverse relationships (adding the opposite) to form a sum and likewise with division problems by using the multiplicative inverse relationships (multiplying by the reciprocal) to form a product.

Students recognize that any order does not apply for expressions mixing addition and multiplication, leading to the need to follow the order of operations.

Classwork


Students complete the table in the Opening Exercise that scaffolds the concept of opposite expressions from the known concept of opposite numbers to find the opposite of the expression 3𝑥𝑥 − 7.

Opening Exercise

Additive inverses have a sum of zero. Fill in the center column of the table with the opposite of the given number or expression, then show the proof that they are opposites. The first row is completed for you.

Expression Opposite Proof of Opposites

𝟏𝟏 −𝟏𝟏 𝟏𝟏 + (−𝟏𝟏) = 𝟎𝟎

𝟑𝟑 −𝟑𝟑 𝟑𝟑 + (−𝟑𝟑) = 𝟎𝟎

−𝟕𝟕 𝟕𝟕 −𝟕𝟕+ 𝟕𝟕 = 𝟎𝟎

−𝟏𝟏𝟐𝟐

𝟏𝟏𝟐𝟐

−𝟏𝟏𝟐𝟐

+𝟏𝟏𝟐𝟐

= 𝟎𝟎

𝒙𝒙 −𝒙𝒙 𝒙𝒙 + (−𝒙𝒙) = 𝟎𝟎

𝟑𝟑𝒙𝒙 −𝟑𝟑𝒙𝒙 𝟑𝟑𝒙𝒙 + (−𝟑𝟑𝒙𝒙) = 𝟎𝟎

𝒙𝒙 + 𝟑𝟑 −𝒙𝒙 + (−𝟑𝟑) (𝒙𝒙 + 𝟑𝟑) + �−𝒙𝒙 + (−𝟑𝟑)� =

�𝒙𝒙 + (−𝒙𝒙)�+ �𝟑𝟑 + (−𝟑𝟑)� = 𝟎𝟎

𝟑𝟑𝒙𝒙 − 𝟕𝟕 −𝟑𝟑𝒙𝒙 + 𝟕𝟕 (𝟑𝟑𝒙𝒙 − 𝟕𝟕) + (−𝟑𝟑𝒙𝒙 + 𝟕𝟕) = 𝟑𝟑𝒙𝒙 + (−𝟕𝟕) + (−𝟑𝟑𝒙𝒙) + 𝟕𝟕 =

�𝟑𝟑𝒙𝒙 + (−𝟑𝟑𝒙𝒙)�+ �(−𝟕𝟕) + 𝟕𝟕� = 𝟎𝟎

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7•3 Lesson 2

Encourage students to provide their answers aloud. When finished, discuss the following:

In the last two rows, explain how the given expression and its opposite compare.

Recall that the opposite of a number, say 𝑎𝑎, satisfies the equation 𝑎𝑎 + (−𝑎𝑎) = 0. We can use this equation to recognize when two expressions are opposites of each other. For example, since (𝑥𝑥 + 3) + �−𝑥𝑥 + (−3)� = 0, we conclude that −𝑥𝑥 + (−3) must be the opposite of 𝑥𝑥 + 3. This is because when either −(𝑥𝑥 + 3) or −𝑥𝑥 + (−3) are substituted into the blank in (𝑥𝑥 + 3) + ________ = 0, the resulting equation is true for every value of 𝑥𝑥. Therefore, the two expressions must be equivalent:

−(𝑥𝑥 + 3) = −𝑥𝑥 + (−3).

Since the opposite of 𝑥𝑥 is −𝑥𝑥 and the opposite of 3 is −3, what can we say about the opposite of the sum of 𝑥𝑥 and 3?

We can say that the opposite of the sum 𝑥𝑥 + 3 is the sum of its opposites (−𝑥𝑥) + (−3). Is this relationship also true for the last example 3𝑥𝑥 − 7?

Yes, because opposites have a sum of zero, so (3𝑥𝑥 − 7) + ___________ = 0. If the expression −3𝑥𝑥 + 7 is substituted in the blank, the resulting equation is true for every value of 𝑥𝑥. The opposite of 3𝑥𝑥 is −3𝑥𝑥, the opposite of (−7) is 7, and the sum of these opposites is −3𝑥𝑥 + 7; therefore, it is true that the opposite of the sum 3𝑥𝑥 + (−7) is the sum of its opposites −3𝑥𝑥 + 7.

�3𝑥𝑥 + (−7)��sum

+ �(−3𝑥𝑥) + 7��opposite

= 0, so −�3𝑥𝑥 + (−7)� = −3𝑥𝑥 + 7.

Can we generalize a rule for the opposite of a sum?

The opposite of a sum is the sum of its opposites.

Tell students that we can use this property as justification for converting the opposites of sums as we work to rewrite expressions in standard form.

Example 1 (5 minutes): Subtracting Expressions

Students and teacher investigate the process for subtracting expressions where the subtrahend is a grouped expression containing two or more terms.

Subtract the expressions in Example 1(a) first by changing subtraction of the expression to adding the expression’s opposite.

Example 1: Subtracting Expressions

a. Subtract: (𝟒𝟒𝟎𝟎+ 𝟗𝟗) − (𝟑𝟑𝟎𝟎+ 𝟐𝟐).

The opposite of a sum is the sum of its opposites. Order of operations

𝟒𝟒𝟎𝟎+ 𝟗𝟗 + �−(𝟑𝟑𝟎𝟎+ 𝟐𝟐)� (𝟒𝟒𝟎𝟎+ 𝟗𝟗) − (𝟑𝟑𝟎𝟎+ 𝟐𝟐) 𝟒𝟒𝟎𝟎+ 𝟗𝟗 + (−𝟑𝟑𝟎𝟎) + (−𝟐𝟐) (𝟒𝟒𝟗𝟗) − (𝟑𝟑𝟐𝟐) 𝟒𝟒𝟗𝟗+ (−𝟑𝟑𝟎𝟎) + (−𝟐𝟐) 𝟏𝟏𝟕𝟕 𝟏𝟏𝟗𝟗+ (−𝟐𝟐) 𝟏𝟏𝟕𝟕

Scaffolding: Finding the opposite (or inverse) of an expression is just like finding the opposite of a mixed number; remember that the opposite of a sum is equal to the sum of its opposites:

−�234� = (−2) + �−

34� ;

−(2 + 𝑥𝑥) = −2 + (−𝑥𝑥).

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7•3 Lesson 2

Next, subtract the expressions using traditional order of operations. Does the difference yield the same number in each case?

Yes. (See previous page.)

Which of the two methods seems more efficient and why?

Answers may vary, but students will likely choose the second method as they are more familiar with it. Which method will have to be used in Example 1(b) and why?

We must add the opposite expression because the terms in parentheses are not like terms, so they cannot be combined as we did with the sum of numbers in Example 1(a).

b. Subtract: (𝟑𝟑𝒙𝒙 + 𝟓𝟓𝟓𝟓 − 𝟒𝟒) − (𝟒𝟒𝒙𝒙 + 𝟏𝟏𝟏𝟏).

𝟑𝟑𝒙𝒙 + 𝟓𝟓𝟓𝟓+ (−𝟒𝟒) + �−(𝟒𝟒𝒙𝒙 + 𝟏𝟏𝟏𝟏)� Subtraction as adding the opposite

𝟑𝟑𝒙𝒙 + 𝟓𝟓𝟓𝟓+ (−𝟒𝟒) + (−𝟒𝟒𝒙𝒙) + (−𝟏𝟏𝟏𝟏) The opposite of a sum is the sum of its opposites.

𝟑𝟑𝒙𝒙 + (−𝟒𝟒𝒙𝒙) + 𝟓𝟓𝟓𝟓 + (−𝟒𝟒) + (−𝟏𝟏𝟏𝟏) Any order, any grouping

−𝒙𝒙 + 𝟓𝟓𝟓𝟓+ (−𝟏𝟏𝟓𝟓) Combining like terms

−𝒙𝒙 + 𝟓𝟓𝟓𝟓 − 𝟏𝟏𝟓𝟓 Subtraction replaces adding the opposite.

Have students check the equivalency of the expressions by substituting 2 for 𝑥𝑥 and 6 for 𝑦𝑦.

(𝟑𝟑𝒙𝒙 + 𝟓𝟓𝟓𝟓 − 𝟒𝟒) − (𝟒𝟒𝒙𝒙 + 𝟏𝟏𝟏𝟏) −𝒙𝒙 + 𝟓𝟓𝟓𝟓 − 𝟏𝟏𝟓𝟓

(𝟑𝟑(𝟐𝟐) + 𝟓𝟓(𝟔𝟔) − 𝟒𝟒) − (𝟒𝟒(𝟐𝟐) + 𝟏𝟏𝟏𝟏) −(𝟐𝟐) + 𝟓𝟓(𝟔𝟔) − 𝟏𝟏𝟓𝟓

(𝟔𝟔+ 𝟑𝟑𝟎𝟎 − 𝟒𝟒) − (𝟖𝟖+ 𝟏𝟏𝟏𝟏) −𝟐𝟐+ 𝟑𝟑𝟎𝟎 + (−𝟏𝟏𝟓𝟓)

(𝟑𝟑𝟔𝟔 − 𝟒𝟒) − (𝟏𝟏𝟗𝟗) 𝟐𝟐𝟖𝟖 + (−𝟏𝟏𝟓𝟓)

𝟑𝟑𝟐𝟐 − 𝟏𝟏𝟗𝟗 𝟏𝟏𝟑𝟑

𝟏𝟏𝟑𝟑

The expressions yield the same number (𝟏𝟏𝟑𝟑) for 𝒙𝒙 = 𝟐𝟐 and 𝟓𝟓 = 𝟔𝟔.

When writing the difference as adding the expression’s opposite in Example 1(b), what happens to the grouped terms that are being subtracted?

When the subtraction is changed to addition, every term in the parentheses that follows must be converted to its opposite.

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7•3 Lesson 2

Example 2 (5 minutes): Combining Expressions Vertically

Students combine expressions by vertically aligning like terms.

Any order, any grouping allows us to write sums and differences as vertical math problems. If we want to combine expressions vertically, we align their like terms vertically.

Example 2: Combining Expressions Vertically

a. Find the sum by aligning the expressions vertically.

(𝟓𝟓𝒂𝒂 + 𝟑𝟑𝒃𝒃 − 𝟔𝟔𝒄𝒄) + (𝟐𝟐𝒂𝒂− 𝟒𝟒𝒃𝒃 + 𝟏𝟏𝟑𝟑𝒄𝒄)

�𝟓𝟓𝒂𝒂 + 𝟑𝟑𝒃𝒃 + (−𝟔𝟔𝒄𝒄)�+ (𝟐𝟐𝒂𝒂 + (−𝟒𝟒𝒃𝒃) + 𝟏𝟏𝟑𝟑𝒄𝒄) Subtraction as adding the opposite

𝟓𝟓𝒂𝒂 + 𝟑𝟑𝒃𝒃 + (−𝟔𝟔𝒄𝒄)+𝟐𝟐𝒂𝒂 + (−𝟒𝟒𝒃𝒃) + 𝟏𝟏𝟑𝟑𝒄𝒄𝟕𝟕𝒂𝒂 + (− 𝒃𝒃) + 𝟕𝟕𝒄𝒄

Align like terms vertically and combine by addition.

𝟕𝟕𝒂𝒂 − 𝒃𝒃 + 𝟕𝟕𝒄𝒄 Adding the opposite is equivalent to subtraction.

b. Find the difference by aligning the expressions vertically.

(𝟐𝟐𝒙𝒙 + 𝟑𝟑𝟓𝟓 − 𝟒𝟒) − (𝟓𝟓𝒙𝒙 + 𝟐𝟐)

�𝟐𝟐𝒙𝒙 + 𝟑𝟑𝟓𝟓+ (−𝟒𝟒)�+ �−𝟓𝟓𝒙𝒙 + (−𝟐𝟐)� Subtraction as adding the opposite

𝟐𝟐𝒙𝒙 + 𝟑𝟑𝟓𝟓+ (−𝟒𝟒)+(−𝟓𝟓𝒙𝒙) + (−𝟐𝟐)−𝟑𝟑𝒙𝒙 + 𝟑𝟑𝟓𝟓+ (−𝟔𝟔)

Align like terms vertically and combine by addition.

−𝟑𝟑𝒙𝒙+ 𝟑𝟑𝟓𝟓 − 𝟔𝟔 Adding the opposite is equivalent to subtraction.

Students should recognize that the subtracted expression in Example 1(b) did not include a term containing the variable 𝑦𝑦, so the 3𝑦𝑦 from the first grouped expression remains unchanged in the answer.

Example 3 (3 minutes): Using Expressions to Solve Problems

Students write an expression representing an unknown real-world value, rewrite it as an equivalent expression, and use the equivalent expression to find the unknown value.

Example 3: Using Expressions to Solve Problems

A stick is 𝒙𝒙 meters long. A string is 𝟒𝟒 times as long as the stick.

a. Express the length of the string in terms of 𝒙𝒙.

The length of the stick in meters is 𝒙𝒙 meters, so the string is 𝟒𝟒 ∙ 𝒙𝒙, or 𝟒𝟒𝒙𝒙, meters long.

b. If the total length of the string and the stick is 𝟏𝟏𝟓𝟓 meters long, how long is the string?

The length of the stick and the string together in meters can be represented by 𝒙𝒙 + 𝟒𝟒𝒙𝒙, or 𝟓𝟓𝒙𝒙. If the length of the stick and string together is 𝟏𝟏𝟓𝟓 meters, the length of the stick is 𝟑𝟑 meters, and the length of the string is 𝟏𝟏𝟐𝟐 meters.

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7•3 Lesson 2

Example 4 (4 minutes): Expressions from Word Problems

Students write expressions described by word problems and rewrite the expressions in standard form.

Example 4: Expressions from Word Problems

It costs Margo a processing fee of $𝟑𝟑 to rent a storage unit, plus $𝟏𝟏𝟕𝟕 per month to keep her belongings in the unit. Her friend Carissa wants to store a box of her belongings in Margo’s storage unit and tells her that she will pay her $𝟏𝟏 toward the processing fee and $𝟑𝟑 for every month that she keeps the box in storage. Write an expression in standard form that represents how much Margo will have to pay for the storage unit if Carissa contributes. Then, determine how much Margo will pay if she uses the storage unit for 𝟔𝟔 months.

Let 𝒎𝒎 represent the number of months that the storage unit is rented.

(𝟏𝟏𝟕𝟕𝒎𝒎+ 𝟑𝟑) − (𝟑𝟑𝒎𝒎+ 𝟏𝟏) Original expression

𝟏𝟏𝟕𝟕𝒎𝒎+ 𝟑𝟑 + �−(𝟑𝟑𝒎𝒎+ 𝟏𝟏)� Subtraction as adding the opposite

𝟏𝟏𝟕𝟕𝒎𝒎+ 𝟑𝟑 + (−𝟑𝟑𝒎𝒎) + (−𝟏𝟏) The opposite of the sum is the sum of its opposites.

𝟏𝟏𝟕𝟕𝒎𝒎+ (−𝟑𝟑𝒎𝒎) + 𝟑𝟑 + (−𝟏𝟏) Any order, any grouping

𝟏𝟏𝟒𝟒𝒎𝒎+ 𝟐𝟐 Combined like terms

This means that Margo will have to pay only $𝟐𝟐 of the processing fee and $𝟏𝟏𝟒𝟒 per month that the storage unit is used.

𝟏𝟏𝟒𝟒(𝟔𝟔) + 𝟐𝟐

𝟖𝟖𝟒𝟒 + 𝟐𝟐

𝟖𝟖𝟔𝟔

Margo will pay $𝟖𝟖𝟔𝟔 toward the storage unit rental for 𝟔𝟔 months of use.

If time allows, encourage students to calculate their answer in other ways and compare their answers.

Example 5 (7 minutes): Extending Use of the Inverse to Division

Students connect the strategy of using the additive inverse to represent a subtraction problem as a sum and using the multiplicative inverse to represent a division problem as a product so that the associative and commutative properties can then be used.

Why do we convert differences into sums using opposites?

The commutative and associative properties do not apply to subtraction; therefore, we convert differences to sums of the opposites so that we can use the any order, any grouping property with addition.

We have seen that the any order, any grouping property can be used with addition or with multiplication. If you consider how we extended the property to subtraction, can we use the any order, any grouping property in a division problem? Explain.

Dividing by a number is equivalent to multiplying by the number’s multiplicative inverse (reciprocal), so division can be converted to multiplication of the reciprocal, similar to how we converted the subtraction of a number to addition using its additive inverse. After converting a quotient to a product, use of the any order, any grouping property is allowed.

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7•3 Lesson 2

Example 5: Extending Use of the Inverse to Division

Multiplicative inverses have a product of 𝟏𝟏. Find the multiplicative inverses of the terms in the first column. Show that the given number and its multiplicative inverse have a product of 𝟏𝟏. Then, use the inverse to write each corresponding expression in standard form. The first row is completed for you.

Given Multiplicative Inverse

Proof—Show that their product is 𝟏𝟏.

Use each inverse to write its corresponding expression below in standard form.

𝟑𝟑 𝟏𝟏𝟑𝟑

𝟑𝟑 ∙𝟏𝟏𝟑𝟑

𝟑𝟑𝟏𝟏∙𝟏𝟏𝟑𝟑

𝟑𝟑𝟑𝟑

𝟏𝟏

𝟏𝟏𝟐𝟐 ÷ 𝟑𝟑

𝟏𝟏𝟐𝟐 ∙𝟏𝟏𝟑𝟑

𝟒𝟒

𝟓𝟓 𝟏𝟏𝟓𝟓

𝟓𝟓 ∙𝟏𝟏𝟓𝟓

𝟓𝟓𝟏𝟏∙𝟏𝟏𝟓𝟓

𝟓𝟓𝟓𝟓

𝟏𝟏

𝟔𝟔𝟓𝟓 ÷ 𝟓𝟓

𝟔𝟔𝟓𝟓 ∙𝟏𝟏𝟓𝟓

𝟏𝟏𝟑𝟑

−𝟐𝟐 −𝟏𝟏𝟐𝟐

−𝟐𝟐 ∙ �−𝟏𝟏𝟐𝟐�

−𝟐𝟐𝟏𝟏∙ �−

𝟏𝟏𝟐𝟐�

𝟐𝟐𝟐𝟐

𝟏𝟏

𝟏𝟏𝟖𝟖÷ (−𝟐𝟐)

𝟏𝟏𝟖𝟖 ∙ �−𝟏𝟏𝟐𝟐�

𝟏𝟏𝟖𝟖 ∙ (−𝟏𝟏) ∙ �𝟏𝟏𝟐𝟐�

−𝟏𝟏𝟖𝟖 ∙𝟏𝟏𝟐𝟐

−𝟗𝟗

−𝟑𝟑𝟓𝟓

−𝟓𝟓𝟑𝟑

−𝟑𝟑𝟓𝟓∙ �−

𝟓𝟓𝟑𝟑�

𝟏𝟏𝟓𝟓𝟏𝟏𝟓𝟓

𝟏𝟏

𝟔𝟔 ÷ �−𝟑𝟑𝟓𝟓�

𝟔𝟔 ∙ �−𝟓𝟓𝟑𝟑�

𝟔𝟔 ∙ (−𝟏𝟏) ∙𝟓𝟓𝟑𝟑

−𝟔𝟔 ∙𝟓𝟓𝟑𝟑

−𝟐𝟐 ∙ 𝟓𝟓 −𝟏𝟏𝟎𝟎

𝒙𝒙 𝟏𝟏𝒙𝒙

𝒙𝒙 ∙𝟏𝟏𝒙𝒙

𝒙𝒙𝟏𝟏∙𝟏𝟏𝒙𝒙

𝒙𝒙𝒙𝒙

𝟏𝟏

𝟓𝟓𝒙𝒙 ÷ 𝒙𝒙

𝟓𝟓𝒙𝒙 ∙𝟏𝟏𝒙𝒙

𝟓𝟓 ∙𝒙𝒙𝒙𝒙

𝟓𝟓 ∙ 𝟏𝟏 𝟓𝟓

𝟐𝟐𝒙𝒙 𝟏𝟏𝟐𝟐𝒙𝒙

𝟐𝟐𝒙𝒙 ∙ �𝟏𝟏𝟐𝟐𝒙𝒙�

𝟐𝟐 ∙ 𝒙𝒙 ∙ �𝟏𝟏𝟐𝟐∙𝟏𝟏𝒙𝒙�

𝟐𝟐 ∙𝟏𝟏𝟐𝟐∙ 𝒙𝒙 ∙

𝟏𝟏𝒙𝒙

𝟏𝟏 ∙ 𝟏𝟏 𝟏𝟏

𝟏𝟏𝟐𝟐𝒙𝒙 ÷ 𝟐𝟐𝒙𝒙

𝟏𝟏𝟐𝟐𝒙𝒙 ∙𝟏𝟏𝟐𝟐𝒙𝒙

𝟏𝟏𝟐𝟐𝒙𝒙𝟐𝟐𝒙𝒙

𝟏𝟏𝟐𝟐𝟐𝟐∙𝒙𝒙𝒙𝒙

𝟔𝟔 ∙ 𝟏𝟏 𝟔𝟔

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7•3 Lesson 2

How do we know that two numbers are multiplicative inverses (reciprocals)?

Recall that the multiplicative inverse of a number, 𝑎𝑎, satisfies the equation 𝑎𝑎 ∙ 1𝑎𝑎 = 1. We can use this

equation to recognize when two expressions are multiplicative inverses of each other.

Since the reciprocal of 𝑥𝑥 is 1𝑥𝑥

, and the reciprocal of 2 is 12

, what can we say about the reciprocal of the product

of 𝑥𝑥 and 2?

We can say that the reciprocal of the product 2𝑥𝑥 is the product of its factor’s reciprocals: 12∙1𝑥𝑥

=12𝑥𝑥

.

What is true about the signs of reciprocals? Why?

The signs of reciprocals are the same because their product must be 1. This can only be obtained when the two numbers in the product have the same sign.

Tell students that because the reciprocal is not complicated by the signs of numbers as in opposites, we can justify converting division to multiplication of the reciprocal by simply stating multiplying by the reciprocal.

Sprint (8 minutes): Generating Equivalent Expressions

Students complete a two-round Sprint exercise (Sprints and answer keys are provided at the end of the lesson.) where they practice their knowledge of combining like terms by addition and/or subtraction. Provide one minute for each round of the Sprint. Refer to the Sprints and Sprint Delivery Script sections in the Module Overview for directions to administer a Sprint. Be sure to provide any answers not completed by the students. (If there is a need for further guided division practice, consider using the division portion of the Problem Set, or other division examples, in place of the provided Sprint exercise.)

Closing (3 minutes)

Why can’t we use any order, any grouping directly with subtraction? With division?

Subtraction and division are not commutative or associative.

How can we use any order, any grouping in expressions where subtraction or division are involved? Subtraction can be rewritten as adding the opposite (additive inverse), and division can be rewritten as

multiplying by the reciprocal (multiplicative inverse).

Relevant Vocabulary

AN EXPRESSION IN EXPANDED FORM: An expression that is written as sums (and/or differences) of products whose factors are numbers, variables, or variables raised to whole number powers is said to be in expanded form. A single number, variable, or a single product of numbers and/or variables is also considered to be in expanded form. Examples of expressions in expanded form include: 𝟑𝟑𝟐𝟐𝟒𝟒, 𝟑𝟑𝒙𝒙, 𝟓𝟓𝒙𝒙 + 𝟑𝟑 − 𝟒𝟒𝟎𝟎, and 𝒙𝒙 + 𝟐𝟐𝒙𝒙 + 𝟑𝟑𝒙𝒙.

TERM: Each summand of an expression in expanded form is called a term. For example, the expression 𝟐𝟐𝒙𝒙 + 𝟑𝟑𝒙𝒙 + 𝟓𝟓 consists of 𝟑𝟑 terms: 𝟐𝟐𝒙𝒙, 𝟑𝟑𝒙𝒙, and 𝟓𝟓.

COEFFICIENT OF THE TERM: The number found by multiplying just the numbers in a term together is called the coefficient. For example, given the product 𝟐𝟐 ∙ 𝒙𝒙 ∙ 𝟒𝟒, its equivalent term is 𝟖𝟖𝒙𝒙. The number 𝟖𝟖 is called the coefficient of the term 𝟖𝟖𝒙𝒙.

AN EXPRESSION IN STANDARD FORM: An expression in expanded form with all of its like terms collected is said to be in standard form. For example, 𝟐𝟐𝟐𝟐+ 𝟑𝟑𝟐𝟐 + 𝟓𝟓 is an expression written in expanded form; however, to be written in standard form, the like terms 𝟐𝟐𝟐𝟐 and 𝟑𝟑𝟐𝟐 must be combined. The equivalent expression 𝟓𝟓𝟐𝟐 + 𝟓𝟓 is written in standard form.

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7•3 Lesson 2


Lesson Summary

Rewrite subtraction as adding the opposite before using any order, any grouping.

Rewrite division as multiplying by the reciprocal before using any order, any grouping.

The opposite of a sum is the sum of its opposites.

Division is equivalent to multiplying by the reciprocal.

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7•3 Lesson 2

Name ___________________________________________________ Date____________________


Exit Ticket 1. Write the expression in standard form.

(4𝑓𝑓 − 3 + 2𝑔𝑔) − (−4𝑔𝑔 + 2)

2. Find the result when 5𝑚𝑚 + 2 is subtracted from 9𝑚𝑚.

3. Write the expression in standard form. 27ℎ ÷ 3ℎ

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7•3 Lesson 2


1. Write the expression in standard form.

(𝟒𝟒𝟒𝟒 − 𝟑𝟑+ 𝟐𝟐𝟐𝟐) − (−𝟒𝟒𝟐𝟐+ 𝟐𝟐)

𝟒𝟒𝟒𝟒 + (−𝟑𝟑) + 𝟐𝟐𝟐𝟐 + �−(−𝟒𝟒𝟐𝟐+ 𝟐𝟐)� Subtraction as adding the opposite

𝟒𝟒𝟒𝟒 + (−𝟑𝟑) + 𝟐𝟐𝟐𝟐 + 𝟒𝟒𝟐𝟐 + (−𝟐𝟐) The opposite of a sum is the sum of its opposites.

𝟒𝟒𝟒𝟒 + 𝟐𝟐𝟐𝟐 + 𝟒𝟒𝟐𝟐 + (−𝟑𝟑) + (−𝟐𝟐) Any order, any grouping

𝟒𝟒𝟒𝟒 + 𝟔𝟔𝟐𝟐 + (−𝟓𝟓) Combined like terms

𝟒𝟒𝟒𝟒 + 𝟔𝟔𝟐𝟐 − 𝟓𝟓 Subtraction as adding the opposite

2. Find the result when 𝟓𝟓𝒎𝒎 + 𝟐𝟐 is subtracted from 𝟗𝟗𝒎𝒎.

𝟗𝟗𝒎𝒎− (𝟓𝟓𝒎𝒎+ 𝟐𝟐) Original expression

𝟗𝟗𝒎𝒎 + �−(𝟓𝟓𝒎𝒎+ 𝟐𝟐)� Subtraction as adding the opposite

𝟗𝟗𝒎𝒎 + (−𝟓𝟓𝒎𝒎) + (−𝟐𝟐) The opposite of a sum is the sum of its opposites.

𝟒𝟒𝒎𝒎 + (−𝟐𝟐) Combined like terms

𝟒𝟒𝒎𝒎− 𝟐𝟐 Subtraction as adding the opposite

3. Write the expression in standard form.

𝟐𝟐𝟕𝟕𝟐𝟐 ÷ 𝟑𝟑𝟐𝟐

𝟐𝟐𝟕𝟕𝟐𝟐 ∙ 𝟏𝟏𝟑𝟑𝟐𝟐 Multiplying by the reciprocal

𝟐𝟐𝟕𝟕𝟐𝟐𝟑𝟑𝟐𝟐

Multiplication

𝟐𝟐𝟕𝟕𝟑𝟑∙𝟐𝟐𝟐𝟐

Any order, any grouping

𝟗𝟗 ∙ 𝟏𝟏

𝟗𝟗

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7•3 Lesson 2


1. Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating each expression using 𝒙𝒙 = 𝟓𝟓.

a. 𝟑𝟑𝒙𝒙 + (𝟐𝟐 − 𝟒𝟒𝒙𝒙)

−𝒙𝒙 + 𝟐𝟐

−𝟓𝟓 + 𝟐𝟐

−𝟑𝟑

𝟑𝟑(𝟓𝟓) + �𝟐𝟐 − 𝟒𝟒(𝟓𝟓)�

𝟏𝟏𝟓𝟓 + �𝟐𝟐 + (−𝟐𝟐𝟎𝟎)�

𝟏𝟏𝟓𝟓 + (−𝟏𝟏𝟖𝟖)

−𝟑𝟑

b. 𝟑𝟑𝒙𝒙 + (−𝟐𝟐+ 𝟒𝟒𝒙𝒙)

𝟕𝟕𝒙𝒙 − 𝟐𝟐

𝟕𝟕(𝟓𝟓) − 𝟐𝟐

𝟑𝟑𝟓𝟓 − 𝟐𝟐

𝟑𝟑𝟑𝟑

𝟑𝟑(𝟓𝟓) + �−𝟐𝟐+ 𝟒𝟒(𝟓𝟓)�

𝟏𝟏𝟓𝟓 + (−𝟐𝟐+ 𝟐𝟐𝟎𝟎)

𝟏𝟏𝟓𝟓 + 𝟏𝟏𝟖𝟖

𝟑𝟑𝟑𝟑

c. −𝟑𝟑𝒙𝒙 + (𝟐𝟐+ 𝟒𝟒𝒙𝒙)

𝒙𝒙 + 𝟐𝟐

𝟓𝟓 + 𝟐𝟐

𝟕𝟕

−𝟑𝟑(𝟓𝟓) + �𝟐𝟐 + 𝟒𝟒(𝟓𝟓)�

−𝟏𝟏𝟓𝟓+ (𝟐𝟐+ 𝟐𝟐𝟎𝟎)

−𝟏𝟏𝟓𝟓+ 𝟐𝟐𝟐𝟐

𝟕𝟕

d. 𝟑𝟑𝒙𝒙 + (−𝟐𝟐− 𝟒𝟒𝒙𝒙)

−𝒙𝒙 − 𝟐𝟐

−𝟓𝟓− 𝟐𝟐

−𝟕𝟕

𝟑𝟑(𝟓𝟓) + �−𝟐𝟐 − 𝟒𝟒(𝟓𝟓)�

𝟏𝟏𝟓𝟓 + �−𝟐𝟐+ �−𝟒𝟒(𝟓𝟓)��

𝟏𝟏𝟓𝟓 + �−𝟐𝟐 + (−𝟐𝟐𝟎𝟎)�

𝟏𝟏𝟓𝟓 + (−𝟐𝟐𝟐𝟐)

−𝟕𝟕

e. 𝟑𝟑𝒙𝒙 − (𝟐𝟐+ 𝟒𝟒𝒙𝒙)

−𝒙𝒙 − 𝟐𝟐

−𝟓𝟓− 𝟐𝟐

−𝟕𝟕

𝟑𝟑(𝟓𝟓) − �𝟐𝟐+ 𝟒𝟒(𝟓𝟓)�

𝟏𝟏𝟓𝟓 − (𝟐𝟐 + 𝟐𝟐𝟎𝟎)

𝟏𝟏𝟓𝟓 − 𝟐𝟐𝟐𝟐

𝟏𝟏𝟓𝟓 + (−𝟐𝟐𝟐𝟐)

−𝟕𝟕

f. 𝟑𝟑𝒙𝒙 − (−𝟐𝟐+ 𝟒𝟒𝒙𝒙)

−𝒙𝒙 + 𝟐𝟐

−𝟓𝟓+ 𝟐𝟐

−𝟑𝟑

𝟑𝟑(𝟓𝟓) − �−𝟐𝟐+ 𝟒𝟒(𝟓𝟓)�

𝟏𝟏𝟓𝟓 − (−𝟐𝟐+ 𝟐𝟐𝟎𝟎)

𝟏𝟏𝟓𝟓 − (𝟏𝟏𝟖𝟖)

𝟏𝟏𝟓𝟓 + (−𝟏𝟏𝟖𝟖)

−𝟑𝟑

g. 𝟑𝟑𝒙𝒙 − (−𝟐𝟐− 𝟒𝟒𝒙𝒙)

𝟕𝟕𝒙𝒙 + 𝟐𝟐

𝟕𝟕(𝟓𝟓) + 𝟐𝟐

𝟑𝟑𝟓𝟓 + 𝟐𝟐

𝟑𝟑𝟕𝟕

𝟑𝟑(𝟓𝟓) − �−𝟐𝟐 − 𝟒𝟒(𝟓𝟓)�

𝟏𝟏𝟓𝟓 − �−𝟐𝟐+ �−𝟒𝟒(𝟓𝟓)��

𝟏𝟏𝟓𝟓 − �−𝟐𝟐 + (−𝟐𝟐𝟎𝟎)�

𝟏𝟏𝟓𝟓 − (−𝟐𝟐𝟐𝟐)

𝟏𝟏𝟓𝟓 + 𝟐𝟐𝟐𝟐

𝟑𝟑𝟕𝟕

h. 𝟑𝟑𝒙𝒙 − (𝟐𝟐 − 𝟒𝟒𝒙𝒙)

𝟕𝟕𝒙𝒙 − 𝟐𝟐

𝟕𝟕(𝟓𝟓) − 𝟐𝟐

𝟑𝟑𝟓𝟓 − 𝟐𝟐

𝟑𝟑𝟑𝟑

𝟑𝟑(𝟓𝟓) − �𝟐𝟐 − 𝟒𝟒(𝟓𝟓)�

𝟏𝟏𝟓𝟓 − �𝟐𝟐 + �−𝟒𝟒(𝟓𝟓)��

𝟏𝟏𝟓𝟓 − �𝟐𝟐 + (−𝟐𝟐𝟎𝟎)�

𝟏𝟏𝟓𝟓 − (−𝟏𝟏𝟖𝟖)

𝟏𝟏𝟓𝟓 + 𝟏𝟏𝟖𝟖

𝟑𝟑𝟑𝟑

i. −𝟑𝟑𝒙𝒙 − (−𝟐𝟐− 𝟒𝟒𝒙𝒙)

𝒙𝒙 + 𝟐𝟐

𝟓𝟓 + 𝟐𝟐

𝟕𝟕

−𝟑𝟑(𝟓𝟓) − �−𝟐𝟐 − 𝟒𝟒(𝟓𝟓)�

−𝟏𝟏𝟓𝟓 − �−𝟐𝟐+ �−𝟒𝟒(𝟓𝟓)��

−𝟏𝟏𝟓𝟓 − �−𝟐𝟐+ (−𝟐𝟐𝟎𝟎)�

−𝟏𝟏𝟓𝟓 − (−𝟐𝟐𝟐𝟐)

−𝟏𝟏𝟓𝟓+ 𝟐𝟐𝟐𝟐

𝟕𝟕

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7•3 Lesson 2

j. In problems (a)–(d) above, what effect does addition have on the terms in parentheses when you removed the parentheses?

By the any grouping property, the terms remained the same with or without the parentheses.

k. In problems (e)–(i), what effect does subtraction have on the terms in parentheses when you removed the parentheses?

The opposite of a sum is the sum of the opposites; each term within the parentheses is changed to its opposite.

2. Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating each expression for the given value of the variable.

a. 𝟒𝟒𝟓𝟓 − (𝟑𝟑 + 𝟓𝟓); 𝟓𝟓 = 𝟐𝟐

𝟑𝟑𝟓𝟓 − 𝟑𝟑

𝟑𝟑(𝟐𝟐) − 𝟑𝟑

𝟔𝟔 − 𝟑𝟑

𝟑𝟑

𝟒𝟒(𝟐𝟐) − (𝟑𝟑+ 𝟐𝟐)

𝟖𝟖 − 𝟓𝟓

𝟖𝟖 + (−𝟓𝟓)

𝟑𝟑

b. (𝟐𝟐𝒃𝒃 + 𝟏𝟏) − 𝒃𝒃; 𝒃𝒃 = −𝟒𝟒

𝒃𝒃 + 𝟏𝟏

−𝟒𝟒+ 𝟏𝟏

−𝟑𝟑

(𝟐𝟐(−𝟒𝟒) + 𝟏𝟏) − (−𝟒𝟒)

(−𝟖𝟖+ 𝟏𝟏) + 𝟒𝟒

(−𝟕𝟕) + 𝟒𝟒

−𝟑𝟑

c. (𝟔𝟔𝒄𝒄 − 𝟒𝟒) − (𝒄𝒄 − 𝟑𝟑); 𝒄𝒄 = −𝟕𝟕

𝟓𝟓𝒄𝒄 − 𝟏𝟏

𝟓𝟓(−𝟕𝟕) − 𝟏𝟏

−𝟑𝟑𝟓𝟓 − 𝟏𝟏

−𝟑𝟑𝟔𝟔

(𝟔𝟔(−𝟕𝟕)− 𝟒𝟒) − (−𝟕𝟕− 𝟑𝟑)

(−𝟒𝟒𝟐𝟐 − 𝟒𝟒) − (−𝟏𝟏𝟎𝟎)

−𝟒𝟒𝟐𝟐+ (−𝟒𝟒) + (𝟏𝟏𝟎𝟎)

−𝟒𝟒𝟔𝟔+ 𝟏𝟏𝟎𝟎

−𝟑𝟑𝟔𝟔

d. (𝒅𝒅+ 𝟑𝟑𝒅𝒅) − (−𝒅𝒅+ 𝟐𝟐); 𝒅𝒅 = 𝟑𝟑

𝟓𝟓𝒅𝒅 − 𝟐𝟐

𝟓𝟓(𝟑𝟑) − 𝟐𝟐

𝟏𝟏𝟓𝟓 − 𝟐𝟐

𝟏𝟏𝟑𝟑

�𝟑𝟑 + 𝟑𝟑(𝟑𝟑)� − (−𝟑𝟑+ 𝟐𝟐)

(𝟑𝟑+ 𝟗𝟗) − (−𝟏𝟏)

𝟏𝟏𝟐𝟐 + 𝟏𝟏

𝟏𝟏𝟑𝟑

e. (−𝟓𝟓𝒙𝒙 − 𝟒𝟒) − (−𝟐𝟐− 𝟓𝟓𝒙𝒙); 𝒙𝒙 = 𝟑𝟑

−𝟐𝟐

(−𝟓𝟓(𝟑𝟑)− 𝟒𝟒) − �−𝟐𝟐 − 𝟓𝟓(𝟑𝟑)�

(−𝟏𝟏𝟓𝟓 − 𝟒𝟒) − (−𝟐𝟐− 𝟏𝟏𝟓𝟓)

(−𝟏𝟏𝟗𝟗) − (−𝟏𝟏𝟕𝟕)

(−𝟏𝟏𝟗𝟗) + 𝟏𝟏𝟕𝟕

−𝟐𝟐

f. 𝟏𝟏𝟏𝟏𝟒𝟒 − (−𝟐𝟐𝟒𝟒+ 𝟐𝟐); 𝟒𝟒 = 𝟏𝟏𝟐𝟐

𝟏𝟏𝟑𝟑𝟒𝟒 − 𝟐𝟐

𝟏𝟏𝟑𝟑�𝟏𝟏𝟐𝟐� − 𝟐𝟐

𝟏𝟏𝟑𝟑𝟐𝟐− 𝟐𝟐

𝟔𝟔𝟏𝟏𝟐𝟐− 𝟐𝟐

𝟒𝟒𝟏𝟏𝟐𝟐

𝟏𝟏𝟏𝟏�𝟏𝟏𝟐𝟐� − �−𝟐𝟐�

𝟏𝟏𝟐𝟐�+ 𝟐𝟐�

𝟏𝟏𝟏𝟏𝟐𝟐− (−𝟏𝟏+ 𝟐𝟐)

𝟏𝟏𝟏𝟏𝟐𝟐− 𝟏𝟏

𝟏𝟏𝟏𝟏𝟐𝟐

+ �−𝟐𝟐𝟐𝟐�

𝟗𝟗𝟐𝟐

𝟒𝟒𝟏𝟏𝟐𝟐

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7•3 Lesson 2

g. −𝟓𝟓𝟐𝟐+ (𝟔𝟔𝟐𝟐− 𝟒𝟒); 𝟐𝟐 = −𝟐𝟐

𝟐𝟐 − 𝟒𝟒

−𝟐𝟐− 𝟒𝟒

−𝟔𝟔

−𝟓𝟓(−𝟐𝟐) + (𝟔𝟔(−𝟐𝟐) − 𝟒𝟒)

𝟏𝟏𝟎𝟎 + (−𝟏𝟏𝟐𝟐 − 𝟒𝟒)

𝟏𝟏𝟎𝟎 + �−𝟏𝟏𝟐𝟐 + (−𝟒𝟒)�

𝟏𝟏𝟎𝟎 + (−𝟏𝟏𝟔𝟔)

−𝟔𝟔

h. (𝟖𝟖𝟐𝟐− 𝟏𝟏) − (𝟐𝟐 + 𝟑𝟑); 𝟐𝟐 = −𝟑𝟑

𝟕𝟕𝟐𝟐 − 𝟒𝟒

𝟕𝟕(−𝟑𝟑) − 𝟒𝟒

−𝟐𝟐𝟏𝟏 − 𝟒𝟒

−𝟐𝟐𝟓𝟓

(𝟖𝟖(−𝟑𝟑)− 𝟏𝟏) − (−𝟑𝟑+ 𝟑𝟑)

(−𝟐𝟐𝟒𝟒 − 𝟏𝟏) − (𝟎𝟎)

(−𝟐𝟐𝟓𝟓) − 𝟎𝟎

−𝟐𝟐𝟓𝟓

i. (𝟕𝟕+ 𝒘𝒘) − (𝒘𝒘 + 𝟕𝟕); 𝒘𝒘 = −𝟒𝟒

𝟎𝟎

�𝟕𝟕 + (−𝟒𝟒)� − (−𝟒𝟒+ 𝟕𝟕)

𝟑𝟑 − 𝟑𝟑

𝟑𝟑 + (−𝟑𝟑)

𝟎𝟎

j. (𝟐𝟐𝟐𝟐+ 𝟗𝟗𝟐𝟐 − 𝟓𝟓) − (𝟔𝟔𝟐𝟐− 𝟒𝟒𝟐𝟐+ 𝟐𝟐); 𝟐𝟐 = −𝟐𝟐 and 𝟐𝟐 = 𝟓𝟓

−𝟒𝟒𝟐𝟐 + 𝟏𝟏𝟑𝟑𝟐𝟐 − 𝟕𝟕 (𝟐𝟐(−𝟐𝟐) + 𝟗𝟗(𝟓𝟓) − 𝟓𝟓) − (𝟔𝟔(−𝟐𝟐) − 𝟒𝟒(𝟓𝟓) + 𝟐𝟐)

−𝟒𝟒(−𝟐𝟐) + 𝟏𝟏𝟑𝟑(𝟓𝟓) − 𝟕𝟕 (−𝟒𝟒+ 𝟒𝟒𝟓𝟓 − 𝟓𝟓) − �−𝟏𝟏𝟐𝟐+ �−𝟒𝟒(𝟓𝟓)�+ 𝟐𝟐�

𝟖𝟖 + 𝟔𝟔𝟓𝟓 + (−𝟕𝟕) (𝟒𝟒𝟏𝟏 − 𝟓𝟓) − (−𝟏𝟏𝟐𝟐+ (−𝟐𝟐𝟎𝟎) + 𝟐𝟐)

𝟕𝟕𝟑𝟑 + (−𝟕𝟕) �𝟒𝟒𝟏𝟏 + (−𝟓𝟓)� − (−𝟑𝟑𝟐𝟐+ 𝟐𝟐)

𝟔𝟔𝟔𝟔 𝟑𝟑𝟔𝟔 − (−𝟑𝟑𝟎𝟎)

𝟑𝟑𝟔𝟔 + 𝟑𝟑𝟎𝟎

𝟔𝟔𝟔𝟔

3. Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating both expressions for the given value of the variable.

a. −𝟑𝟑(𝟖𝟖𝒙𝒙); 𝒙𝒙 = 𝟏𝟏𝟒𝟒

−𝟐𝟐𝟒𝟒𝒙𝒙

−𝟐𝟐𝟒𝟒�𝟏𝟏𝟒𝟒�

−𝟐𝟐𝟒𝟒𝟒𝟒

−𝟔𝟔

−𝟑𝟑�𝟖𝟖�𝟏𝟏𝟒𝟒��

−𝟑𝟑(𝟐𝟐)

−𝟔𝟔

b. 𝟓𝟓 ∙ 𝒌𝒌 ∙ (−𝟕𝟕); 𝒌𝒌 = 𝟑𝟑𝟓𝟓

−𝟑𝟑𝟓𝟓𝒌𝒌

−𝟑𝟑𝟓𝟓�𝟑𝟑𝟓𝟓�

−𝟏𝟏𝟎𝟎𝟓𝟓𝟓𝟓

−𝟐𝟐𝟏𝟏

𝟓𝟓�𝟑𝟑𝟓𝟓� (−𝟕𝟕)

𝟑𝟑(−𝟕𝟕)

−𝟐𝟐𝟏𝟏

c. 𝟐𝟐(−𝟔𝟔𝒙𝒙) ∙ 𝟐𝟐; 𝒙𝒙 = 𝟑𝟑𝟒𝟒

−𝟐𝟐𝟒𝟒𝒙𝒙

−𝟐𝟐𝟒𝟒�𝟑𝟑𝟒𝟒�

−𝟕𝟕𝟐𝟐𝟒𝟒

−𝟏𝟏𝟖𝟖

𝟐𝟐�−𝟔𝟔�𝟑𝟑𝟒𝟒�� ∙ 𝟐𝟐

𝟐𝟐�−𝟑𝟑�𝟑𝟑𝟐𝟐�� ∙ 𝟐𝟐

𝟐𝟐(−𝟑𝟑) �𝟑𝟑𝟐𝟐� (𝟐𝟐)

−𝟔𝟔(𝟑𝟑)

−𝟏𝟏𝟖𝟖

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7•3 Lesson 2

d. −𝟑𝟑(𝟖𝟖𝒙𝒙) + 𝟔𝟔(𝟒𝟒𝒙𝒙); 𝒙𝒙 = 𝟐𝟐

𝟎𝟎

−𝟑𝟑�𝟖𝟖(𝟐𝟐)�+ 𝟔𝟔�𝟒𝟒(𝟐𝟐)�

−𝟑𝟑(𝟏𝟏𝟔𝟔) + 𝟔𝟔(𝟖𝟖)

−𝟒𝟒𝟖𝟖+ 𝟒𝟒𝟖𝟖

𝟎𝟎

e. 𝟖𝟖(𝟓𝟓𝒎𝒎) + 𝟐𝟐(𝟑𝟑𝒎𝒎); 𝒎𝒎 = −𝟐𝟐

𝟒𝟒𝟔𝟔𝒎𝒎

𝟒𝟒𝟔𝟔(−𝟐𝟐)

−𝟗𝟗𝟐𝟐

𝟖𝟖�𝟓𝟓(−𝟐𝟐)�+ 𝟐𝟐�𝟑𝟑(−𝟐𝟐)�

𝟖𝟖(−𝟏𝟏𝟎𝟎) + 𝟐𝟐(−𝟔𝟔)

−𝟖𝟖𝟎𝟎+ (−𝟏𝟏𝟐𝟐)

−𝟗𝟗𝟐𝟐

f. −𝟔𝟔(𝟐𝟐𝟐𝟐) + 𝟑𝟑𝒂𝒂(𝟑𝟑); 𝟐𝟐 = 𝟏𝟏𝟑𝟑;

𝒂𝒂 = 𝟐𝟐𝟑𝟑

−𝟏𝟏𝟐𝟐𝟐𝟐+ 𝟗𝟗𝒂𝒂

−𝟏𝟏𝟐𝟐�𝟏𝟏𝟑𝟑�+ 𝟗𝟗�

𝟐𝟐𝟑𝟑�

−𝟏𝟏𝟐𝟐𝟑𝟑

+𝟏𝟏𝟖𝟖𝟑𝟑

−𝟒𝟒+ 𝟔𝟔

𝟐𝟐

−𝟔𝟔�𝟐𝟐�𝟏𝟏𝟑𝟑�� + 𝟑𝟑�

𝟐𝟐𝟑𝟑� (𝟑𝟑)

−𝟔𝟔�𝟐𝟐𝟑𝟑�+ 𝟐𝟐(𝟑𝟑)

−𝟒𝟒+ 𝟔𝟔

𝟐𝟐

4. Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating both expressions for the given value of the variable.

a. 𝟖𝟖𝒙𝒙 ÷ 𝟐𝟐; 𝒙𝒙 = −𝟏𝟏𝟒𝟒

𝟒𝟒𝒙𝒙

𝟒𝟒�−𝟏𝟏𝟒𝟒�

−𝟏𝟏

𝟖𝟖�−𝟏𝟏𝟒𝟒�÷ 𝟐𝟐

−𝟐𝟐÷ 𝟐𝟐

−𝟏𝟏

b. 𝟏𝟏𝟖𝟖𝒘𝒘 ÷ 𝟔𝟔; 𝒘𝒘 = 𝟔𝟔

𝟑𝟑𝒘𝒘

𝟑𝟑(𝟔𝟔)

𝟏𝟏𝟖𝟖

𝟏𝟏𝟖𝟖(𝟔𝟔) ÷ 𝟔𝟔

𝟏𝟏𝟎𝟎𝟖𝟖÷ 𝟔𝟔

𝟏𝟏𝟖𝟖

c. 𝟐𝟐𝟓𝟓𝟐𝟐÷ 𝟓𝟓𝟐𝟐; 𝟐𝟐 = −𝟐𝟐

𝟓𝟓

𝟐𝟐𝟓𝟓(−𝟐𝟐) ÷ �𝟓𝟓(−𝟐𝟐)�

−𝟓𝟓𝟎𝟎÷ (−𝟏𝟏𝟎𝟎)

𝟓𝟓

d. 𝟑𝟑𝟑𝟑𝟓𝟓÷ 𝟏𝟏𝟏𝟏𝟓𝟓; 𝟓𝟓 = −𝟐𝟐

𝟑𝟑

𝟑𝟑𝟑𝟑(−𝟐𝟐) ÷ �𝟏𝟏𝟏𝟏(−𝟐𝟐)� (−𝟔𝟔𝟔𝟔) ÷ (−𝟐𝟐𝟐𝟐)

𝟑𝟑

e. 𝟓𝟓𝟔𝟔𝒌𝒌÷ 𝟐𝟐𝒌𝒌; 𝒌𝒌 = 𝟑𝟑

𝟐𝟐𝟖𝟖

𝟓𝟓𝟔𝟔(𝟑𝟑) ÷ �𝟐𝟐(𝟑𝟑)�

𝟏𝟏𝟔𝟔𝟖𝟖÷ 𝟔𝟔

𝟐𝟐𝟖𝟖

f. 𝟐𝟐𝟒𝟒𝒙𝒙𝟓𝟓÷ 𝟔𝟔𝟓𝟓; 𝒙𝒙 = −𝟐𝟐;𝟓𝟓 = 𝟑𝟑

𝟒𝟒𝒙𝒙

𝟒𝟒(−𝟐𝟐)

−𝟖𝟖

𝟐𝟐𝟒𝟒(−𝟐𝟐)(𝟑𝟑) ÷ �𝟔𝟔(𝟑𝟑)�

−𝟒𝟒𝟖𝟖(𝟑𝟑) ÷ 𝟏𝟏𝟖𝟖

−𝟏𝟏𝟒𝟒𝟒𝟒÷ 𝟏𝟏𝟖𝟖

−𝟖𝟖

5. For each problem (a)–(g), write an expression in standard form.

a. Find the sum of −𝟑𝟑𝒙𝒙 and 𝟖𝟖𝒙𝒙.

−𝟑𝟑𝒙𝒙 + 𝟖𝟖𝒙𝒙

𝟓𝟓𝒙𝒙

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7•3 Lesson 2

b. Find the sum of −𝟕𝟕𝟐𝟐 and 𝟒𝟒𝟐𝟐 + 𝟐𝟐.

−𝟕𝟕𝟐𝟐+ (𝟒𝟒𝟐𝟐 + 𝟐𝟐)

−𝟑𝟑𝟐𝟐+ 𝟐𝟐

c. Find the difference when 𝟔𝟔𝟐𝟐 is subtracted from 𝟐𝟐𝟐𝟐− 𝟒𝟒.

(𝟐𝟐𝟐𝟐− 𝟒𝟒) − 𝟔𝟔𝟐𝟐

−𝟒𝟒𝟐𝟐− 𝟒𝟒

d. Find the difference when −𝟑𝟑𝟑𝟑 − 𝟕𝟕 is subtracted from 𝟑𝟑 + 𝟒𝟒.

(𝟑𝟑+ 𝟒𝟒) − (−𝟑𝟑𝟑𝟑− 𝟕𝟕)

𝟒𝟒𝟑𝟑 + 𝟏𝟏𝟏𝟏

e. Find the result when 𝟏𝟏𝟑𝟑𝟐𝟐 + 𝟐𝟐 is subtracted from 𝟏𝟏𝟏𝟏 + 𝟓𝟓𝟐𝟐.

(𝟏𝟏𝟏𝟏+ 𝟓𝟓𝟐𝟐) − (𝟏𝟏𝟑𝟑𝟐𝟐+ 𝟐𝟐)

−𝟖𝟖𝟐𝟐+ 𝟗𝟗

f. Find the result when −𝟏𝟏𝟖𝟖𝒎𝒎− 𝟒𝟒 is added to 𝟒𝟒𝒎𝒎− 𝟏𝟏𝟒𝟒.

(𝟒𝟒𝒎𝒎− 𝟏𝟏𝟒𝟒) + (−𝟏𝟏𝟖𝟖𝒎𝒎− 𝟒𝟒)

−𝟏𝟏𝟒𝟒𝒎𝒎− 𝟏𝟏𝟖𝟖

g. What is the result when −𝟐𝟐𝒙𝒙 + 𝟗𝟗 is taken away from −𝟕𝟕𝒙𝒙 + 𝟐𝟐?

(−𝟕𝟕𝒙𝒙+ 𝟐𝟐) − (−𝟐𝟐𝒙𝒙 + 𝟗𝟗)

−𝟓𝟓𝒙𝒙 − 𝟕𝟕

6. Marty and Stewart are stuffing envelopes with index cards. They are putting 𝒙𝒙 index cards in each envelope. When they are finished, Marty has 𝟏𝟏𝟓𝟓 stuffed envelopes and 𝟒𝟒 extra index cards, and Stewart has 𝟏𝟏𝟐𝟐 stuffed envelopes and 𝟔𝟔 extra index cards. Write an expression in standard form that represents the number of index cards the boys started with. Explain what your expression means.

They inserted the same number of index cards in each envelope, but that number is unknown, 𝒙𝒙. An expression that represents Marty’s index cards is 𝟏𝟏𝟓𝟓𝒙𝒙 + 𝟒𝟒 because he had 𝟏𝟏𝟓𝟓 envelopes and 𝟒𝟒 cards left over. An expression that represents Stewart’s index cards is 𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟔𝟔 because he had 𝟏𝟏𝟐𝟐 envelopes and 𝟔𝟔 left over cards. Their total number of cards together would be:

𝟏𝟏𝟓𝟓𝒙𝒙 + 𝟒𝟒 + 𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟔𝟔

𝟏𝟏𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟒𝟒 + 𝟔𝟔 𝟐𝟐𝟕𝟕𝒙𝒙 + 𝟏𝟏𝟎𝟎

This means that altogether, they have 𝟐𝟐𝟕𝟕 envelopes with 𝒙𝒙 index cards in each, plus another 𝟏𝟏𝟎𝟎 leftover index cards.

A STORY OF RATIOS


42






7•3 Lesson 2

7. The area of the pictured rectangle below is 𝟐𝟐𝟒𝟒𝒃𝒃 𝐟𝐟𝐭𝐭𝟐𝟐. Its width is 𝟐𝟐𝒃𝒃 𝐟𝐟𝐭𝐭. Find the height of the rectangle and name any properties used with the appropriate step.

𝟐𝟐𝟒𝟒𝒃𝒃 ÷ 𝟐𝟐𝒃𝒃

𝟐𝟐𝟒𝟒𝒃𝒃 ∙ 𝟏𝟏𝟐𝟐𝒃𝒃 Multiplying the reciprocal

𝟐𝟐𝟒𝟒𝒃𝒃𝟐𝟐𝒃𝒃

Multiplication

𝟐𝟐𝟒𝟒𝟐𝟐∙𝒃𝒃𝒃𝒃

Any order, any grouping in multiplication

𝟏𝟏𝟐𝟐 ∙ 𝟏𝟏 𝟏𝟏𝟐𝟐

The height of the rectangle is 𝟏𝟏𝟐𝟐 𝐟𝐟𝐭𝐭.

𝟐𝟐𝒃𝒃 𝐟𝐟𝐭𝐭.

𝟐𝟐𝟒𝟒𝒃𝒃 𝐟𝐟𝐭𝐭𝟐𝟐 ___ 𝐟𝐟𝐭𝐭.

A STORY OF RATIOS


43






7•3 Lesson 2

Generating Equivalent Expressions—Round 1 Directions: Write each as an equivalent expression in standard form as quickly and as accurately as possible within the allotted time.

1. 1 + 1 23. 4𝑥𝑥 + 6𝑥𝑥 − 12𝑥𝑥

2. 1 + 1 + 1 24. 4𝑥𝑥 − 6𝑥𝑥 + 4𝑥𝑥

3. (1 + 1) + 1 25. 7𝑥𝑥 − 2𝑥𝑥 + 3

4. (1 + 1) + (1 + 1) 26. (4𝑥𝑥 + 3) + 𝑥𝑥

5. (1 + 1) + (1 + 1 + 1) 27. (4𝑥𝑥 + 3) + 2𝑥𝑥

6. 𝑥𝑥 + 𝑥𝑥 28. (4𝑥𝑥 + 3) + 3𝑥𝑥

7. 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 29. (4𝑥𝑥 + 3) + 5𝑥𝑥

8. (𝑥𝑥 + 𝑥𝑥) + 𝑥𝑥 30. (4𝑥𝑥 + 3) + 6𝑥𝑥

9. (𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥) 31. (11𝑥𝑥 + 2) − 2

10. (𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) 32. (11𝑥𝑥 + 2) − 3

11. (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) 33. (11𝑥𝑥 + 2) − 4

12. 2𝑥𝑥 + 𝑥𝑥 34. (11𝑥𝑥 + 2) − 7

13. 3𝑥𝑥 + 𝑥𝑥 35. (3𝑥𝑥 − 9) + (3𝑥𝑥 + 5)

14. 4𝑥𝑥 + 𝑥𝑥 36. (11 − 5𝑥𝑥) + (4𝑥𝑥 + 2)

15. 7𝑥𝑥 + 𝑥𝑥 37. (2𝑥𝑥 + 3𝑦𝑦) + (4𝑥𝑥 + 𝑦𝑦)

16. 7𝑥𝑥 + 2𝑥𝑥 38. (5𝑥𝑥 + 1.3𝑦𝑦) + (2.9𝑥𝑥 − 0.6𝑦𝑦)

17. 7𝑥𝑥 + 3𝑥𝑥 39. (2.6𝑥𝑥 − 4.8𝑦𝑦) + (6.5𝑥𝑥 − 1.1𝑦𝑦)

18. 10𝑥𝑥 − 𝑥𝑥 40. �34𝑥𝑥 −

12𝑦𝑦� + �−

74𝑥𝑥 −

52𝑦𝑦�

19. 10𝑥𝑥 − 5𝑥𝑥 41. �−25𝑥𝑥 −

79𝑦𝑦� + �−

710

𝑥𝑥 −23𝑦𝑦�

20. 10𝑥𝑥 − 10𝑥𝑥 42. �12𝑥𝑥 −

14𝑦𝑦� + �−

35𝑥𝑥 +

56𝑦𝑦�

21. 10𝑥𝑥 − 11𝑥𝑥 43. �1.2𝑥𝑥 −34𝑦𝑦� − �−

35𝑥𝑥 + 2.25𝑦𝑦�

22. 10𝑥𝑥 − 12𝑥𝑥 44. (3.375𝑥𝑥 − 8.9𝑦𝑦) − �−758𝑥𝑥 − 5

25𝑦𝑦�

Number Correct: ______

A STORY OF RATIOS


44






7•3 Lesson 2

Generating Equivalent Expressions—Round 1 [KEY] Directions: Write each as an equivalent expression in standard form as quickly and as accurately as possible within the allotted time.

1. 1 + 1 𝟐𝟐 23. 4𝑥𝑥 + 6𝑥𝑥 − 12𝑥𝑥 −𝟐𝟐𝒙𝒙

2. 1 + 1 + 1 𝟑𝟑 24. 4𝑥𝑥 − 6𝑥𝑥 + 4𝑥𝑥 𝟐𝟐𝒙𝒙

3. (1 + 1) + 1 𝟑𝟑 25. 7𝑥𝑥 − 2𝑥𝑥 + 3 𝟓𝟓𝒙𝒙 + 𝟑𝟑

4. (1 + 1) + (1 + 1) 𝟒𝟒 26. (4𝑥𝑥 + 3) + 𝑥𝑥 𝟓𝟓𝒙𝒙 + 𝟑𝟑

5. (1 + 1) + (1 + 1 + 1) 𝟓𝟓 27. (4𝑥𝑥 + 3) + 2𝑥𝑥 𝟔𝟔𝒙𝒙 + 𝟑𝟑

6. 𝑥𝑥 + 𝑥𝑥 𝟐𝟐𝒙𝒙 28. (4𝑥𝑥 + 3) + 3𝑥𝑥 𝟕𝟕𝒙𝒙 + 𝟑𝟑

7. 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 𝟑𝟑𝒙𝒙 29. (4𝑥𝑥 + 3) + 5𝑥𝑥 𝟗𝟗𝒙𝒙 + 𝟑𝟑

8. (𝑥𝑥 + 𝑥𝑥) + 𝑥𝑥 𝟑𝟑𝒙𝒙 30. (4𝑥𝑥 + 3) + 6𝑥𝑥 𝟏𝟏𝟎𝟎𝒙𝒙 + 𝟑𝟑

9. (𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥) 𝟒𝟒𝒙𝒙 31. (11𝑥𝑥 + 2) − 2 𝟏𝟏𝟏𝟏𝒙𝒙

10. (𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) 𝟓𝟓𝒙𝒙 32. (11𝑥𝑥 + 2) − 3 𝟏𝟏𝟏𝟏𝒙𝒙 − 𝟏𝟏

11. (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) 𝟔𝟔𝒙𝒙 33. (11𝑥𝑥 + 2) − 4 𝟏𝟏𝟏𝟏𝒙𝒙 − 𝟐𝟐

12. 2𝑥𝑥 + 𝑥𝑥 𝟑𝟑𝒙𝒙 34. (11𝑥𝑥 + 2) − 7 𝟏𝟏𝟏𝟏𝒙𝒙 − 𝟓𝟓

13. 3𝑥𝑥 + 𝑥𝑥 𝟒𝟒𝒙𝒙 35. (3𝑥𝑥 − 9) + (3𝑥𝑥 + 5) 𝟔𝟔𝒙𝒙 − 𝟒𝟒

14. 4𝑥𝑥 + 𝑥𝑥 𝟓𝟓𝒙𝒙 36. (11 − 5𝑥𝑥) + (4𝑥𝑥 + 2) 𝟏𝟏𝟑𝟑 − 𝒙𝒙 or − 𝒙𝒙 + 𝟏𝟏𝟑𝟑

15. 7𝑥𝑥 + 𝑥𝑥 𝟖𝟖𝒙𝒙 37. (2𝑥𝑥 + 3𝑦𝑦) + (4𝑥𝑥 + 𝑦𝑦) 𝟔𝟔𝒙𝒙 + 𝟒𝟒𝟓𝟓

16. 7𝑥𝑥 + 2𝑥𝑥 𝟗𝟗𝒙𝒙 38. (5𝑥𝑥 + 1.3𝑦𝑦) + (2.9𝑥𝑥 − 0.6𝑦𝑦) 𝟕𝟕.𝟗𝟗𝒙𝒙 + 𝟎𝟎.𝟕𝟕𝟓𝟓

17. 7𝑥𝑥 + 3𝑥𝑥 𝟏𝟏𝟎𝟎𝒙𝒙 39. (2.6𝑥𝑥 − 4.8𝑦𝑦) + (6.5𝑥𝑥 − 1.1𝑦𝑦) 𝟗𝟗.𝟏𝟏𝒙𝒙 − 𝟓𝟓.𝟗𝟗𝟓𝟓

18. 10𝑥𝑥 − 𝑥𝑥 𝟗𝟗𝒙𝒙 40. �34𝑥𝑥 −

12𝑦𝑦� + �−

74𝑥𝑥 −

52𝑦𝑦� −𝒙𝒙 − 𝟑𝟑𝟓𝟓

19. 10𝑥𝑥 − 5𝑥𝑥 𝟓𝟓𝒙𝒙 41. �−25𝑥𝑥 −

79𝑦𝑦� + �−

710

𝑥𝑥 −23𝑦𝑦� −

𝟏𝟏𝟏𝟏𝟏𝟏𝟎𝟎

𝒙𝒙 −𝟏𝟏𝟑𝟑𝟗𝟗𝟓𝟓

20. 10𝑥𝑥 − 10𝑥𝑥 𝟎𝟎 42. �12𝑥𝑥 −

14𝑦𝑦� + �−

35𝑥𝑥 +

56𝑦𝑦� −

𝟏𝟏𝟏𝟏𝟎𝟎

𝒙𝒙 +𝟕𝟕𝟏𝟏𝟐𝟐

𝟓𝟓

21. 10𝑥𝑥 − 11𝑥𝑥 −𝟏𝟏𝒙𝒙 or − 𝒙𝒙 43. �1.2𝑥𝑥 −34𝑦𝑦� − �−

35𝑥𝑥 + 2.25𝑦𝑦�

𝟗𝟗𝟓𝟓𝒙𝒙 − 𝟑𝟑𝟓𝟓

22. 10𝑥𝑥 − 12𝑥𝑥 −𝟐𝟐𝒙𝒙 44. (3.375𝑥𝑥 − 8.9𝑦𝑦) − �−758𝑥𝑥 − 5

25𝑦𝑦� 𝟏𝟏𝟏𝟏𝒙𝒙 −

𝟕𝟕𝟐𝟐𝟓𝟓

A STORY OF RATIOS


45






7•3 Lesson 2

Generating Equivalent Expressions—Round 2 Directions: Write each as an equivalent expression in standard form as quickly and as accurately as possible within the allotted time.

1. 1 + 1 + 1 23. 3𝑥𝑥 + 5𝑥𝑥 − 4𝑥𝑥

2. 1 + 1 + 1 + 1 24. 8𝑥𝑥 − 6𝑥𝑥 + 4𝑥𝑥

3. (1 + 1 + 1) + 1 25. 7𝑥𝑥 − 4𝑥𝑥 + 5

4. (1 + 1 + 1) + (1 + 1) 26. (9𝑥𝑥 − 1) + 𝑥𝑥

5. (1 + 1 + 1) + (1 + 1 + 1) 27. (9𝑥𝑥 − 1) + 2𝑥𝑥

6. 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 28. (9𝑥𝑥 − 1) + 3𝑥𝑥

7. 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 29. (9𝑥𝑥 − 1) + 5𝑥𝑥

8. (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) + 𝑥𝑥 30. (9𝑥𝑥 − 1) + 6𝑥𝑥

9. (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥) 31. (−3𝑥𝑥 + 3) − 2

10. (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) 32. (−3𝑥𝑥 + 3) − 3

11. (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥) 33. (−3𝑥𝑥 + 3) − 4

12. 𝑥𝑥 + 2𝑥𝑥 34. (−3𝑥𝑥 + 3) − 5

13. 𝑥𝑥 + 4𝑥𝑥 35. (5𝑥𝑥 − 2) + (2𝑥𝑥 + 5)

14. 𝑥𝑥 + 6𝑥𝑥 36. (8 − 𝑥𝑥) + (3𝑥𝑥 + 2)

15. 𝑥𝑥 + 8𝑥𝑥 37. (5𝑥𝑥 + 𝑦𝑦) + (𝑥𝑥 + 𝑦𝑦)

16. 7𝑥𝑥 + 𝑥𝑥 38. �52𝑥𝑥 +

32𝑦𝑦� + �

112𝑥𝑥 −

34𝑦𝑦�

17. 8𝑥𝑥 + 2𝑥𝑥 39. �16𝑥𝑥 −

38𝑦𝑦� + �

23𝑥𝑥 −

74𝑦𝑦�

18. 2𝑥𝑥 − 𝑥𝑥 40. (9.7𝑥𝑥 − 3.8𝑦𝑦) + (−2.8𝑥𝑥 + 4.5𝑦𝑦)

19. 2𝑥𝑥 − 2𝑥𝑥 41. (1.65𝑥𝑥 − 2.73𝑦𝑦) + (−1.35𝑥𝑥 + 3.76𝑦𝑦)

20. 2𝑥𝑥 − 3𝑥𝑥 42. (6.51𝑥𝑥 − 4.39𝑦𝑦) + (−7.46𝑥𝑥 + 8.11𝑦𝑦)

21. 2𝑥𝑥 − 4𝑥𝑥 43. �0.7𝑥𝑥 −29𝑦𝑦� − �−

75𝑥𝑥 + 2

13𝑦𝑦�

22. 2𝑥𝑥 − 8𝑥𝑥 44. (8.4𝑥𝑥 − 2.25𝑦𝑦) − �−212𝑥𝑥 − 4

38𝑦𝑦�

Number Correct: ______ Improvement: ______

A STORY OF RATIOS


46






7•3 Lesson 2

Generating Equivalent Expressions—Round 2 [KEY] Directions: Write each as an equivalent expression in standard form as quickly and as accurately as possible within the allotted time.

1. 1 + 1 + 1 𝟑𝟑 23. 3𝑥𝑥 + 5𝑥𝑥 − 4𝑥𝑥 𝟒𝟒𝒙𝒙

2. 1 + 1 + 1 + 1 𝟒𝟒 24. 8𝑥𝑥 − 6𝑥𝑥 + 4𝑥𝑥 𝟔𝟔𝒙𝒙

3. (1 + 1 + 1) + 1 𝟒𝟒 25. 7𝑥𝑥 − 4𝑥𝑥 + 5 𝟑𝟑𝒙𝒙 + 𝟓𝟓

4. (1 + 1 + 1) + (1 + 1) 𝟓𝟓 26. (9𝑥𝑥 − 1) + 𝑥𝑥 𝟏𝟏𝟎𝟎𝒙𝒙 − 𝟏𝟏

5. (1 + 1 + 1) + (1 + 1 + 1) 𝟔𝟔 27. (9𝑥𝑥 − 1) + 2𝑥𝑥 𝟏𝟏𝟏𝟏𝒙𝒙 − 𝟏𝟏

6. 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 𝟑𝟑𝒙𝒙 28. (9𝑥𝑥 − 1) + 3𝑥𝑥 𝟏𝟏𝟐𝟐𝒙𝒙 − 𝟏𝟏

7. 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 𝟒𝟒𝒙𝒙 29. (9𝑥𝑥 − 1) + 5𝑥𝑥 𝟏𝟏𝟒𝟒𝒙𝒙 − 𝟏𝟏

8. (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) + 𝑥𝑥 𝟒𝟒𝒙𝒙 30. (9𝑥𝑥 − 1) + 6𝑥𝑥 𝟏𝟏𝟓𝟓𝒙𝒙 − 𝟏𝟏

9. (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥) 𝟓𝟓𝒙𝒙 31. (−3𝑥𝑥 + 3) − 2 −𝟑𝟑𝒙𝒙 + 𝟏𝟏

10. (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) 𝟔𝟔𝒙𝒙 32. (−3𝑥𝑥 + 3) − 3 −𝟑𝟑𝒙𝒙

11. (𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥) + (𝑥𝑥 + 𝑥𝑥) 𝟔𝟔𝒙𝒙 33. (−3𝑥𝑥 + 3) − 4 −𝟑𝟑𝒙𝒙 − 𝟏𝟏

12. 𝑥𝑥 + 2𝑥𝑥 𝟑𝟑𝒙𝒙 34. (−3𝑥𝑥 + 3) − 5 −𝟑𝟑𝒙𝒙 − 𝟐𝟐

13. 𝑥𝑥 + 4𝑥𝑥 𝟓𝟓𝒙𝒙 35. (5𝑥𝑥 − 2) + (2𝑥𝑥 + 5) 𝟕𝟕𝒙𝒙 + 𝟑𝟑

14. 𝑥𝑥 + 6𝑥𝑥 𝟕𝟕𝒙𝒙 36. (8 − 𝑥𝑥) + (3𝑥𝑥 + 2) 𝟏𝟏𝟎𝟎 + 𝟐𝟐𝒙𝒙

15. 𝑥𝑥 + 8𝑥𝑥 𝟗𝟗𝒙𝒙 37. (5𝑥𝑥 + 𝑦𝑦) + (𝑥𝑥 + 𝑦𝑦) 𝟔𝟔𝒙𝒙 + 𝟐𝟐𝟓𝟓

16. 7𝑥𝑥 + 𝑥𝑥 𝟖𝟖𝒙𝒙 38. �52𝑥𝑥 +

32𝑦𝑦� + �

112𝑥𝑥 −

34𝑦𝑦� 𝟖𝟖𝒙𝒙 +

𝟑𝟑𝟒𝟒𝟓𝟓

17. 8𝑥𝑥 + 2𝑥𝑥 𝟏𝟏𝟎𝟎𝒙𝒙 39. �16𝑥𝑥 −

38𝑦𝑦� + �

23𝑥𝑥 −

74𝑦𝑦�

𝟓𝟓𝟔𝟔𝒙𝒙 −

𝟏𝟏𝟕𝟕𝟖𝟖𝟓𝟓

18. 2𝑥𝑥 − 𝑥𝑥 𝒙𝒙 or 𝟏𝟏𝒙𝒙 40. (9.7𝑥𝑥 − 3.8𝑦𝑦) + (−2.8𝑥𝑥 + 4.5𝑦𝑦) 𝟔𝟔.𝟗𝟗𝒙𝒙 + 𝟎𝟎.𝟕𝟕𝟓𝟓

19. 2𝑥𝑥 − 2𝑥𝑥 𝟎𝟎 41. (1.65𝑥𝑥 − 2.73𝑦𝑦) + (−1.35𝑥𝑥 + 3.76𝑦𝑦) 𝟎𝟎.𝟑𝟑𝒙𝒙 + 𝟏𝟏.𝟎𝟎𝟑𝟑𝟓𝟓

20. 2𝑥𝑥 − 3𝑥𝑥 −𝒙𝒙 or − 𝟏𝟏𝒙𝒙 42. (6.51𝑥𝑥 − 4.39𝑦𝑦) + (−7.46𝑥𝑥 + 8.11𝑦𝑦) −𝟎𝟎.𝟗𝟗𝟓𝟓𝒙𝒙+ 𝟑𝟑.𝟕𝟕𝟐𝟐𝟓𝟓

21. 2𝑥𝑥 − 4𝑥𝑥 −𝟐𝟐𝒙𝒙 43. �0.7𝑥𝑥 −29𝑦𝑦� − �−

75𝑥𝑥 + 2

13𝑦𝑦�

𝟐𝟐𝟏𝟏𝟏𝟏𝟎𝟎

𝒙𝒙 −𝟐𝟐𝟑𝟑𝟗𝟗𝟓𝟓

22. 2𝑥𝑥 − 8𝑥𝑥 −𝟔𝟔𝒙𝒙 44. (8.4𝑥𝑥 − 2.25𝑦𝑦) − �−212𝑥𝑥 − 4

38𝑦𝑦�

𝟏𝟏𝟎𝟎𝟗𝟗𝟏𝟏𝟎𝟎

𝒙𝒙 +𝟏𝟏𝟕𝟕𝟖𝟖𝟓𝟓

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Lesson 3: Writing Products as Sums and Sums as Products

7•3 Lesson 3


Student Outcomes

Students use area and rectangular array models and the distributive property to write products as sums and sums as products.

Students use the fact that the opposite of a number is the same as multiplying by −1 to write the opposite of a sum in standard form.

Students recognize that rewriting an expression in a different form can shed light on the problem and how the quantities in it are related.

Classwork


Students create tape diagrams to represent the problem and solution.

Opening Exercise

Solve the problem using a tape diagram. A sum of money was shared between George and Benjamin in a ratio of 𝟑𝟑 ∶ 𝟒𝟒. If the sum of money was $𝟓𝟓𝟓𝟓.𝟎𝟎𝟎𝟎, how much did George get?

Have students label one unit as 𝑥𝑥 in the diagram.

What does the rectangle labeled 𝑥𝑥 represent? $8.00

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7•3 Lesson 3

Example 1 (3 minutes)

Example 1

Represent 𝟑𝟑 + 𝟐𝟐 using a tape diagram.

Represent it also in this fashion:

Now, let’s represent another expression, 𝑥𝑥 + 2. Make sure the units are the same size when you are drawing the known 2 units.

Represent 𝒙𝒙 + 𝟐𝟐 using a tape diagram.

Note the size of the units that represent 2 in the expression 𝑥𝑥 + 2. Using the size of these units, can you predict what value 𝑥𝑥 represents?

Approximately six units

Draw a rectangular array for 𝟑𝟑(𝟑𝟑 + 𝟐𝟐).

Then, have students draw a similar array for 3(𝑥𝑥 + 2).

Draw an array for 𝟑𝟑(𝒙𝒙 + 𝟐𝟐).

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7•3 Lesson 3

Key Terms

DISTRIBUTIVE PROPERTY: The distributive property can be written as the identity

𝒂𝒂(𝒃𝒃+ 𝒄𝒄) = 𝒂𝒂𝒃𝒃 + 𝒂𝒂𝒄𝒄 for all numbers 𝒂𝒂, 𝒃𝒃, and 𝒄𝒄.

Determine the area of the shaded region. 6

Determine the area of the unshaded region. 3𝑥𝑥

Record the areas of each region:

Introduce the term distributive property in the Key Terms box from the Student Materials.

Exercise 1 (3 minutes) Exercise 1

Determine the area of each region using the distributive property.

Answers: 𝟏𝟏𝟏𝟏𝟓𝟓, 𝟓𝟓𝟓𝟓

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7•3 Lesson 3

Draw in the units in the diagram for students.

Is it easier to just imagine the 176 and 55 square units?

Yes


Model the creation of the tape diagrams for the following expressions. Students draw the tape diagrams on the student pages and use the models for discussion.

Example 2

Draw a tape diagram to represent each expression.

a. (𝒙𝒙 + 𝒚𝒚) + (𝒙𝒙 + 𝒚𝒚) + (𝒙𝒙 + 𝒚𝒚)

b. (𝒙𝒙 + 𝒙𝒙 + 𝒙𝒙) + (𝒚𝒚+ 𝒚𝒚+ 𝒚𝒚)

c. 𝟑𝟑𝒙𝒙 + 𝟑𝟑𝒚𝒚

Or

d. 𝟑𝟑(𝒙𝒙 + 𝒚𝒚)

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7•3 Lesson 3

Ask students to explain to their neighbors why all of these expressions are equivalent.

Discuss how to rearrange the units representing 𝑥𝑥 and 𝑦𝑦 into each of the configurations on the previous page.

What can we conclude about all of these expressions?

They are all equivalent.

How does 3(𝑥𝑥 + 𝑦𝑦) = 3𝑥𝑥 + 3𝑦𝑦? Three groups of (𝑥𝑥 + 𝑦𝑦) is the same as multiplying 3 with the 𝑥𝑥 and the 𝑦𝑦.

How do you know the three representations of the expressions are equivalent?

The arithmetic, algebraic, and graphic representations are equivalent. Problem (c) is the standard form of problems (b) and (d). Problem (a) is the equivalent of problems (b) and (c) before the distributive property is applied. Problem (b) is the expanded form before collecting like terms.

Under which conditions would each representation be most useful?

Either 3(𝑥𝑥 + 𝑦𝑦) or 3𝑥𝑥 + 3𝑦𝑦 because it is clear to see that there are 3 groups of (𝑥𝑥 + 𝑦𝑦), which is the product of the sum of 𝑥𝑥 and 𝑦𝑦, or that the second expression is the sum of 3𝑥𝑥 and 3𝑦𝑦.

Which model best represents the distributive property?

Summarize the distributive property.


Example 3

Find an equivalent expression by modeling with a rectangular array and applying the distributive property to the expression 𝟓𝟓(𝟖𝟖𝒙𝒙 + 𝟑𝟑).

Distribute the factor to all the terms. 𝟓𝟓 (𝟖𝟖𝒙𝒙+ 𝟑𝟑)

Multiply . 𝟓𝟓(𝟖𝟖𝒙𝒙) + 𝟓𝟓(𝟑𝟑)

𝟒𝟒𝟎𝟎𝒙𝒙 + 𝟏𝟏𝟓𝟓

Substitute given numerical value to demonstrate equivalency. Let 𝒙𝒙 = 𝟐𝟐

𝟓𝟓(𝟖𝟖𝒙𝒙 + 𝟑𝟑) = 𝟓𝟓 (𝟖𝟖(𝟐𝟐) + 𝟑𝟑) = 𝟓𝟓(𝟏𝟏𝟓𝟓+ 𝟑𝟑) = 𝟓𝟓(𝟏𝟏𝟏𝟏) = 𝟏𝟏𝟓𝟓 𝟒𝟒𝟎𝟎𝒙𝒙 + 𝟏𝟏𝟓𝟓 = 𝟒𝟒𝟎𝟎(𝟐𝟐) + 𝟏𝟏𝟓𝟓 = 𝟖𝟖𝟎𝟎 + 𝟏𝟏𝟓𝟓 = 𝟏𝟏𝟓𝟓

Both equal 𝟏𝟏𝟓𝟓, so the expressions are equal.

Scaffolding: For the struggling student, draw a rectangular array for 5(3). The number of squares in the rectangular array is the product because the factors are 5 and 3. Therefore, 5(3) = 15 is represented.

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7•3 Lesson 3

Exercise 2 (3 minutes)

Allow students to work on the problems independently and share aloud their equivalent expressions. Substitute numerical values to demonstrate equivalency.

Exercise 2

For parts (a) and (b), draw an array for each expression and apply the distributive property to expand each expression. Substitute the given numerical values to demonstrate equivalency.

a. 𝟐𝟐(𝒙𝒙 + 𝟏𝟏), 𝒙𝒙 = 𝟓𝟓

𝟐𝟐𝒙𝒙 + 𝟐𝟐, 𝟏𝟏𝟐𝟐

b. 𝟏𝟏𝟎𝟎(𝟐𝟐𝒄𝒄 + 𝟓𝟓), 𝒄𝒄 = 𝟏𝟏

𝟐𝟐𝟎𝟎𝒄𝒄 + 𝟓𝟓𝟎𝟎, 𝟏𝟏𝟎𝟎

For parts (c) and (d), apply the distributive property. Substitute the given numerical values to demonstrate equivalency.

c. 𝟑𝟑(𝟒𝟒𝟒𝟒 − 𝟏𝟏), 𝟒𝟒 = 𝟐𝟐

𝟏𝟏𝟐𝟐𝟒𝟒 − 𝟑𝟑, 𝟐𝟐𝟏𝟏

d. 𝟏𝟏(−𝟑𝟑𝟑𝟑 − 𝟏𝟏𝟏𝟏), 𝟑𝟑 = 𝟏𝟏𝟎𝟎

−𝟐𝟐𝟏𝟏𝟑𝟑 − 𝟏𝟏𝟏𝟏, −𝟑𝟑𝟓𝟓𝟏𝟏


Example 4

Rewrite the expression (𝟓𝟓𝒙𝒙+ 𝟏𝟏𝟓𝟓) ÷ 𝟑𝟑 in standard form using the distributive property.

(𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟓𝟓) ×𝟏𝟏𝟑𝟑

(𝟓𝟓𝒙𝒙)𝟏𝟏𝟑𝟑

+ (𝟏𝟏𝟓𝟓)𝟏𝟏𝟑𝟑

𝟐𝟐𝒙𝒙 + 𝟓𝟓

How can we rewrite the expression so that the distributive property can be used?

We can change from dividing by 3 to multiplying by 13

.

𝟐𝟐𝒄𝒄 𝟓𝟓

𝟏𝟏𝟎𝟎

𝟐𝟐𝟎𝟎𝒄𝒄

𝟓𝟓𝟎𝟎

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7•3 Lesson 3


Rewrite the expressions in standard form.

a. (𝟐𝟐𝒃𝒃 + 𝟏𝟏𝟐𝟐) ÷ 𝟐𝟐

𝟏𝟏𝟐𝟐

(𝟐𝟐𝒃𝒃 + 𝟏𝟏𝟐𝟐)

𝟏𝟏𝟐𝟐

(𝟐𝟐𝒃𝒃) +𝟏𝟏𝟐𝟐

(𝟏𝟏𝟐𝟐)

𝒃𝒃 + 𝟓𝟓

b. (𝟐𝟐𝟎𝟎𝟑𝟑 − 𝟖𝟖) ÷ 𝟒𝟒

𝟏𝟏𝟒𝟒

(𝟐𝟐𝟎𝟎𝟑𝟑 − 𝟖𝟖)

𝟏𝟏𝟒𝟒

(𝟐𝟐𝟎𝟎𝟑𝟑) −𝟏𝟏𝟒𝟒

(𝟖𝟖)

𝟓𝟓𝟑𝟑 − 𝟐𝟐

c. (𝟒𝟒𝟏𝟏𝒈𝒈− 𝟏𝟏) ÷ 𝟏𝟏

𝟏𝟏𝟏𝟏

(𝟒𝟒𝟏𝟏𝒈𝒈− 𝟏𝟏)

𝟏𝟏𝟏𝟏

(𝟒𝟒𝟏𝟏𝒈𝒈) −𝟏𝟏𝟏𝟏

(𝟏𝟏)

𝟏𝟏𝒈𝒈 − 𝟏𝟏


Model the following exercise with the use of rectangular arrays. Discuss:

What is a verbal explanation of 4(𝑥𝑥 + 𝑦𝑦 + 𝑧𝑧)?

There are 4 groups of the sum of 𝑥𝑥, 𝑦𝑦, and 𝑧𝑧.

Example 5

Expand the expression 𝟒𝟒(𝒙𝒙 + 𝒚𝒚 + 𝒛𝒛).

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7•3 Lesson 3


Instruct students to complete the exercise individually.

Exercise 4

Expand the expression from a product to a sum by removing grouping symbols using an area model and the repeated use of the distributive property: 𝟑𝟑(𝒙𝒙 + 𝟐𝟐𝒚𝒚+ 𝟓𝟓𝒛𝒛).

Repeated use of the distributive property: Visually:

𝟑𝟑(𝒙𝒙 + 𝟐𝟐𝒚𝒚 + 𝟓𝟓𝒛𝒛)

𝟑𝟑 ∙ 𝒙𝒙 + 𝟑𝟑 ∙ 𝟐𝟐𝒚𝒚+ 𝟑𝟑 ∙ 𝟓𝟓𝒛𝒛

𝟑𝟑𝒙𝒙 + 𝟑𝟑 ∙ 𝟐𝟐 ∙ 𝒚𝒚+ 𝟑𝟑 ∙ 𝟓𝟓 ∙ 𝒛𝒛

𝟑𝟑𝒙𝒙 + 𝟓𝟓𝒚𝒚+ 𝟏𝟏𝟓𝟓𝒛𝒛


After reading the problem aloud with the class, use different lengths to represent 𝑠𝑠 in order to come up with expressions with numerical values.

Example 6

A square fountain area with side length 𝒔𝒔 𝐟𝐟𝐟𝐟. is bordered by a single row of square tiles as shown. Express the total number of tiles needed in terms of 𝒔𝒔 three different ways.

What if 𝑠𝑠 = 4? How many tiles would you need to border the fountain?

I would need 20 tiles to border the fountain—four for each side and one for each corner.

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7•3 Lesson 3

What if 𝑠𝑠 = 2? How many tiles would you need to border the fountain?

I would need 12 tiles to border the fountain—two for each side and one for each corner.

What pattern or generalization do you notice?

Answers may vary. Sample response: There is one tile for each corner and four times the number of tiles to fit one side length.

After using numerical values, allow students two minutes to create as many expressions as they can think of to find the total number of tiles in the border in terms of 𝑠𝑠. Reconvene by asking students to share their expressions with the class from their seat.

Which expressions would you use and why?

Although all the expressions are equivalent, 4(𝑠𝑠 + 1), or 4𝑠𝑠 + 4, is useful because it is the most simplified, concise form. It is in standard form with all like terms collected.

Sample Responses:

𝒔𝒔 + 𝒔𝒔 + 𝒔𝒔 + 𝒔𝒔 + 𝟒𝟒. Explanation: There are 𝟒𝟒 sides of 𝒔𝒔 tiles and 𝟒𝟒 extra tiles for the corners.

𝟒𝟒(𝒔𝒔 + 𝟏𝟏) Explanation: There are four groups of 𝒔𝒔 tiles plus 𝟏𝟏 corner tile.

𝟐𝟐𝒔𝒔 + 𝟐𝟐(𝒔𝒔 + 𝟐𝟐) Explanation: There are 𝟐𝟐 opposite sides of 𝒔𝒔 tiles plus 𝟐𝟐 groups of a side of 𝒔𝒔 tiles plus 𝟐𝟐 corner tiles.

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7•3 Lesson 3

Closing (3 minutes)

What are some of the methods used to write products as sums?

We used the distributive property and rectangular arrays. In terms of a rectangular array and equivalent expressions, what does the product form represent, and what

does the sum form represent?

The total area represents the expression written in sum form, and the length and width represent the expressions written in product form.


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7•3 Lesson 3

Name ___________________________________________________ Date____________________


Exit Ticket A square fountain area with side length 𝑠𝑠 ft. is bordered by two rows of square tiles along its perimeter as shown. Express the total number of grey tiles (the second border of tiles) needed in terms of 𝑠𝑠 three different ways.

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7•3 Lesson 3

or 𝒔𝒔 + 𝒔𝒔 + 𝒔𝒔 + 𝒔𝒔 + 𝟏𝟏𝟐𝟐


A square fountain area with side length 𝒔𝒔 𝐟𝐟𝐟𝐟. is bordered by two rows of square tiles along its perimeter as shown. Express the total number of grey tiles (the second border of tiles) needed in terms of 𝒔𝒔 three different ways.

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7•3 Lesson 3


1.

a. Write two equivalent expressions that represent the rectangular array below.

𝟑𝟑(𝟐𝟐𝒂𝒂+ 𝟓𝟓) or 𝟓𝟓𝒂𝒂 + 𝟏𝟏𝟓𝟓

b. Verify informally that the two expressions are equivalent using substitution.

Let 𝒂𝒂 = 𝟒𝟒.

𝟑𝟑(𝟐𝟐𝒂𝒂 + 𝟓𝟓)

𝟑𝟑(𝟐𝟐(𝟒𝟒) + 𝟓𝟓)

𝟑𝟑(𝟖𝟖+ 𝟓𝟓)

𝟑𝟑(𝟏𝟏𝟑𝟑)

𝟑𝟑𝟏𝟏

𝟓𝟓𝒂𝒂 + 𝟏𝟏𝟓𝟓

𝟓𝟓(𝟒𝟒) + 𝟏𝟏𝟓𝟓

𝟐𝟐𝟒𝟒+ 𝟏𝟏𝟓𝟓

𝟑𝟑𝟏𝟏

2. You and your friend made up a basketball shooting game. Every shot made from the free throw line is worth 𝟑𝟑 points, and every shot made from the half-court mark is worth 𝟓𝟓 points. Write an equation that represents the total number of points, 𝑷𝑷, if 𝟒𝟒 represents the number of shots made from the free throw line, and 𝒉𝒉 represents the number of shots made from half-court. Explain the equation in words.

𝑷𝑷 = 𝟑𝟑𝟒𝟒 + 𝟓𝟓𝒉𝒉 or 𝑷𝑷 = 𝟑𝟑(𝟒𝟒 + 𝟐𝟐𝒉𝒉)

The total number of points can be determined by multiplying each free throw shot by 𝟑𝟑 and then adding that to the product of each half-court shot multiplied by 𝟓𝟓.

The total number of points can also be determined by adding the number of free throw shots to twice the number of half-court shots and then multiplying the sum by three.

3. Use a rectangular array to write the products in standard form.

a. 𝟐𝟐(𝒙𝒙 + 𝟏𝟏𝟎𝟎)

𝟐𝟐𝒙𝒙 + 𝟐𝟐𝟎𝟎

b. 𝟑𝟑(𝟒𝟒𝒃𝒃 + 𝟏𝟏𝟐𝟐𝒄𝒄 + 𝟏𝟏𝟏𝟏)

𝟏𝟏𝟐𝟐𝒃𝒃 + 𝟑𝟑𝟓𝟓𝒄𝒄 + 𝟑𝟑𝟑𝟑

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7•3 Lesson 3

4. Use the distributive property to write the products in standard form.

a. 𝟑𝟑(𝟐𝟐𝒙𝒙 − 𝟏𝟏)

𝟓𝟓𝒙𝒙 − 𝟑𝟑

b. 𝟏𝟏𝟎𝟎(𝒃𝒃 + 𝟒𝟒𝒄𝒄)

𝟏𝟏𝟎𝟎𝒃𝒃 + 𝟒𝟒𝟎𝟎𝒄𝒄

c. 𝟏𝟏(𝒈𝒈− 𝟓𝟓𝒉𝒉)

𝟏𝟏𝒈𝒈 − 𝟒𝟒𝟓𝟓𝒉𝒉

d. 𝟏𝟏(𝟒𝟒𝟒𝟒− 𝟓𝟓𝟓𝟓− 𝟐𝟐)

𝟐𝟐𝟖𝟖𝟒𝟒 − 𝟑𝟑𝟓𝟓𝟓𝟓− 𝟏𝟏𝟒𝟒

e. 𝒂𝒂(𝒃𝒃+ 𝒄𝒄 + 𝟏𝟏)

𝒂𝒂𝒃𝒃 + 𝒂𝒂𝒄𝒄 + 𝒂𝒂

f. (𝟖𝟖𝒋𝒋 − 𝟑𝟑𝒍𝒍 + 𝟏𝟏)𝟓𝟓

𝟒𝟒𝟖𝟖𝒋𝒋 − 𝟏𝟏𝟖𝟖𝒍𝒍 + 𝟓𝟓𝟒𝟒

g. (𝟒𝟒𝟎𝟎𝒔𝒔 + 𝟏𝟏𝟎𝟎𝟎𝟎𝒕𝒕) ÷ 𝟏𝟏𝟎𝟎

𝟒𝟒𝒔𝒔 + 𝟏𝟏𝟎𝟎𝒕𝒕

h. (𝟒𝟒𝟖𝟖𝒑𝒑 + 𝟐𝟐𝟒𝟒) ÷ 𝟓𝟓

𝟖𝟖𝒑𝒑 + 𝟒𝟒

i. (𝟐𝟐𝒃𝒃 + 𝟏𝟏𝟐𝟐) ÷ 𝟐𝟐

𝒃𝒃 + 𝟓𝟓

j. (𝟐𝟐𝟎𝟎𝟑𝟑 − 𝟖𝟖) ÷ 𝟒𝟒

𝟓𝟓𝟑𝟑 − 𝟐𝟐

k. (𝟒𝟒𝟏𝟏𝒈𝒈− 𝟏𝟏) ÷ 𝟏𝟏

𝟏𝟏𝒈𝒈 − 𝟏𝟏

l. (𝟏𝟏𝟒𝟒𝒈𝒈+ 𝟐𝟐𝟐𝟐𝒉𝒉) ÷ 𝟏𝟏𝟐𝟐

𝟐𝟐𝟖𝟖𝒈𝒈 + 𝟒𝟒𝟒𝟒𝒉𝒉

5. Write the expression in standard form by expanding and collecting like terms.

a. 𝟒𝟒(𝟖𝟖𝟓𝟓− 𝟏𝟏𝟒𝟒) + 𝟓𝟓(𝟑𝟑𝟒𝟒− 𝟒𝟒𝟓𝟓)

𝟖𝟖𝟓𝟓− 𝟏𝟏𝟎𝟎𝟒𝟒

b. 𝟏𝟏(𝟑𝟑 − 𝒔𝒔) + 𝟓𝟓(𝟐𝟐𝟑𝟑 − 𝟐𝟐𝒔𝒔)

𝟏𝟏𝟏𝟏𝟑𝟑 − 𝟏𝟏𝟏𝟏𝒔𝒔

c. 𝟏𝟏𝟐𝟐(𝟏𝟏 − 𝟑𝟑𝒈𝒈) + 𝟖𝟖(𝒈𝒈+ 𝟒𝟒)

−𝟐𝟐𝟖𝟖𝒈𝒈+ 𝟖𝟖𝟒𝟒 + 𝟏𝟏𝟐𝟐

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7•3 Lesson 4


Student Outcomes

Students use an area model to write products as sums and sums as products. Students use the fact that the opposite of a number is the same as multiplying by −1 to write the opposite of a

sum in standard form.

Students recognize that rewriting an expression in a different form can shed light on the problem and how the quantities in it are related.

Classwork


Give students two minutes to write equivalent expressions using the distributive property for the first four problems. Then, ask students to try to factor out a common factor and write equivalent expressions for the last four problems.

Example 1

a. 𝟐𝟐(𝒙𝒙 + 𝟓𝟓) 𝟐𝟐𝒙𝒙 + 𝟏𝟏𝟏𝟏

b. 𝟑𝟑(𝒙𝒙 + 𝟒𝟒) 𝟑𝟑𝒙𝒙 + 𝟏𝟏𝟐𝟐

c. 𝟔𝟔(𝒙𝒙 + 𝟏𝟏) 𝟔𝟔𝒙𝒙 + 𝟔𝟔

d. 𝟕𝟕(𝒙𝒙 − 𝟑𝟑) 𝟕𝟕𝒙𝒙 − 𝟐𝟐𝟏𝟏

e. 𝟓𝟓(𝒙𝒙 + 𝟔𝟔) 𝟓𝟓𝒙𝒙 + 𝟑𝟑𝟏𝟏

f. 𝟖𝟖(𝒙𝒙 + 𝟏𝟏) 𝟖𝟖𝒙𝒙 + 𝟖𝟖

g. 𝟑𝟑(𝒙𝒙 − 𝟒𝟒) 𝟑𝟑𝒙𝒙 − 𝟏𝟏𝟐𝟐

h. 𝟓𝟓(𝟑𝟑𝒙𝒙 + 𝟒𝟒) 𝟏𝟏𝟓𝟓𝒙𝒙 + 𝟐𝟐𝟏𝟏

What is happening when you factor and write equivalent expressions for parts (e), (f), (g), and (h)?

In the same way that dividing is the opposite or inverse operation of multiplying, factoring is the opposite of expanding.

What are the terms being divided by?

They are being divided by a common factor.

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7•3 Lesson 4

Have students write an expression that is equivalent to 8𝑥𝑥 + 4.

Would it be incorrect to factor out a 2 instead of a 4?

It would not be incorrect, but in order to factor completely, we would need to factor out another 2.

8𝑥𝑥 + 4 Commutative property

4(2𝑥𝑥) + 4(1) Equivalent expression

4(2𝑥𝑥 + 1) Distributive property


Students work independently and share their answers with a partner. Discuss together as a class.

Exercise 1

Rewrite the expressions as a product of two factors.

a. 𝟕𝟕𝟐𝟐𝒕𝒕+ 𝟖𝟖

𝟖𝟖(𝟗𝟗𝒕𝒕+ 𝟏𝟏)

b. 𝟓𝟓𝟓𝟓𝒂𝒂 + 𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏(𝟓𝟓𝒂𝒂+ 𝟏𝟏)

c. 𝟑𝟑𝟔𝟔𝒛𝒛 + 𝟕𝟕𝟐𝟐

𝟑𝟑𝟔𝟔(𝒛𝒛 + 𝟐𝟐)

d. 𝟏𝟏𝟒𝟒𝟒𝟒𝒒𝒒 − 𝟏𝟏𝟓𝟓

𝟑𝟑(𝟒𝟒𝟖𝟖𝒒𝒒− 𝟓𝟓)

e. 𝟑𝟑𝒓𝒓+ 𝟑𝟑𝒔𝒔

𝟑𝟑(𝒓𝒓+ 𝒔𝒔)


In this example, let the variables 𝑥𝑥 and 𝑦𝑦 stand for positive integers, and let 2𝑥𝑥, 12𝑦𝑦, and 8 represent the area of three rectangles. The goal is to find the lengths of each individual region with areas of 2𝑥𝑥, 12𝑦𝑦, and 8, so that all three regions have the same width. Let students explore different possibilities.

Example 2

Let the variables 𝒙𝒙 and 𝒚𝒚 stand for positive integers, and let 𝟐𝟐𝒙𝒙, 𝟏𝟏𝟐𝟐𝒚𝒚, and 𝟖𝟖 represent the area of three regions in the array. Determine the length and width of each rectangle if the width is the same for each rectangle.

What does 2𝑥𝑥 represent in the first region of the array?

The region has an area of 2𝑥𝑥 or can be covered by 2𝑥𝑥 square units.

What does the entire array represent?

The entire array represents 2𝑥𝑥 + 12𝑦𝑦 + 8 square units.

𝟏𝟏𝟐𝟐𝒚𝒚 𝟖𝟖

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7•3 Lesson 4

𝟐𝟐

𝒙𝒙 𝟔𝟔𝒚𝒚 𝟒𝟒

What is the common factor of 2𝑥𝑥, 12𝑦𝑦, and 8?

The common factor of 2𝑥𝑥, 12𝑦𝑦, and 8 is 2.

What are the missing values, and how do you know?

The missing values are 𝑥𝑥, 6𝑦𝑦, and 4. If the products are given in the area of the regions, divide the regions by 2 to determine the missing values.

Write the expression as a sum and then as a product of two factors. 2𝑥𝑥 + 12𝑦𝑦 + 8 2(𝑥𝑥 + 6𝑦𝑦 + 4)

How does this exercise differ from the exercises we did during the previous lesson? How is this exercise similar to the ones we did during the previous lesson?

We are doing the inverse of writing products as sums. Before, we wrote a product as a sum using the distributive property. Now, we are writing a sum as a product using the distributive property.


Have students work on the following exercise individually and discuss the results as a class.

Exercise 2

a. Write the product and sum of the expressions being represented in the rectangular array.

𝟐𝟐(𝟏𝟏𝟐𝟐𝟏𝟏+ 𝟒𝟒𝟒𝟒 + 𝟑𝟑), 𝟐𝟐𝟒𝟒𝟏𝟏 + 𝟖𝟖𝟒𝟒 + 𝟔𝟔

b. Factor 𝟒𝟒𝟖𝟖𝟒𝟒+ 𝟔𝟔𝟏𝟏𝟔𝟔+ 𝟐𝟐𝟒𝟒 by finding the greatest common factor of the terms.

𝟏𝟏𝟐𝟐(𝟒𝟒𝟒𝟒+ 𝟓𝟓𝟔𝟔+ 𝟐𝟐)

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7•3 Lesson 4


For each expression, write each sum as a product of two factors. Emphasize the importance of the distributive property. Use various equivalent expressions to justify equivalency.

a. 𝟐𝟐 ∙ 𝟑𝟑 + 𝟓𝟓 ∙ 𝟑𝟑

Both have a common factor of 𝟑𝟑, so the two factors would be 𝟑𝟑(𝟐𝟐+ 𝟓𝟓). Demonstrate that 𝟑𝟑(𝟕𝟕) is equivalent to 𝟔𝟔 + 𝟏𝟏𝟓𝟓, or 𝟐𝟐𝟏𝟏.

b. (𝟐𝟐+ 𝟓𝟓) + (𝟐𝟐+ 𝟓𝟓) + (𝟐𝟐+ 𝟓𝟓)

This expression is 𝟑𝟑 groups of (𝟐𝟐+ 𝟓𝟓) or 𝟑𝟑(𝟐𝟐) + 𝟑𝟑(𝟓𝟓), which is 𝟑𝟑(𝟐𝟐+ 𝟓𝟓).

c. 𝟐𝟐 ∙ 𝟐𝟐 + (𝟓𝟓 + 𝟐𝟐) + (𝟓𝟓 ∙ 𝟐𝟐)

Rewrite the expression as 𝟐𝟐 ∙ 𝟐𝟐 + (𝟓𝟓 ∙ 𝟐𝟐) + (𝟐𝟐+ 𝟓𝟓), so 𝟐𝟐(𝟐𝟐+ 𝟓𝟓) + (𝟐𝟐+ 𝟓𝟓), which equals 𝟑𝟑(𝟐𝟐+ 𝟓𝟓).

d. 𝒙𝒙 ∙ 𝟑𝟑 + 𝟓𝟓 ∙ 𝟑𝟑

The greatest common factor is 𝟑𝟑, so factor out the 𝟑𝟑: 𝟑𝟑(𝒙𝒙 + 𝟓𝟓).

e. (𝒙𝒙 + 𝟓𝟓) + (𝒙𝒙 + 𝟓𝟓) + (𝒙𝒙 + 𝟓𝟓)

Similar to part (b), this is 𝟑𝟑 groups of (𝒙𝒙 + 𝟓𝟓), so 𝟑𝟑(𝒙𝒙 + 𝟓𝟓).

f. 𝟐𝟐𝒙𝒙 + (𝟓𝟓+ 𝒙𝒙) + 𝟓𝟓 ∙ 𝟐𝟐

Combine like terms and then identify the common factor. 𝟑𝟑𝒙𝒙 + 𝟏𝟏𝟓𝟓, where 𝟑𝟑 is the common factor: 𝟑𝟑(𝒙𝒙 + 𝟓𝟓). Or,

𝟐𝟐𝒙𝒙 + 𝟐𝟐 ∙ 𝟓𝟓+ (𝒙𝒙 + 𝟓𝟓), so that 𝟐𝟐(𝒙𝒙 + 𝟓𝟓) + (𝒙𝒙 + 𝟓𝟓) = 𝟑𝟑(𝒙𝒙 + 𝟓𝟓). Or, use the associative property and write:

𝟐𝟐𝒙𝒙 + (𝟓𝟓 ⋅ 𝟐𝟐) + (𝟓𝟓+ 𝒙𝒙)

𝟐𝟐(𝒙𝒙 + 𝟓𝟓) + (𝟓𝟓 + 𝒙𝒙)

𝟑𝟑(𝒙𝒙 + 𝟓𝟓).

g. 𝒙𝒙 ∙ 𝟑𝟑 + 𝒚𝒚 ∙ 𝟑𝟑

The greatest common factor is 𝟑𝟑, so 𝟑𝟑(𝒙𝒙+ 𝒚𝒚).

h. (𝒙𝒙 + 𝒚𝒚) + (𝒙𝒙 + 𝒚𝒚) + (𝒙𝒙 + 𝒚𝒚)

There are 𝟑𝟑 groups of (𝒙𝒙 + 𝒚𝒚), so 𝟑𝟑(𝒙𝒙 + 𝒚𝒚).

i. 𝟐𝟐𝒙𝒙 + (𝒚𝒚+ 𝒙𝒙) + 𝟐𝟐𝒚𝒚

Combine like terms and then identify the common factor. 𝟑𝟑𝒙𝒙 + 𝟑𝟑𝒚𝒚, where 𝟑𝟑 is the common factor. 𝟑𝟑(𝒙𝒙 + 𝒚𝒚). Or, 𝟐𝟐𝒙𝒙 + 𝟐𝟐𝒚𝒚+ (𝒙𝒙 + 𝒚𝒚), so that 𝟐𝟐(𝒙𝒙 + 𝒚𝒚) + (𝒙𝒙 + 𝒚𝒚) is equivalent to 𝟑𝟑(𝒙𝒙 + 𝒚𝒚). Or, use the associative property, and write:

𝟐𝟐𝒙𝒙 + 𝟐𝟐𝒚𝒚+ (𝒚𝒚+ 𝒙𝒙)

𝟐𝟐(𝒙𝒙 + 𝒚𝒚) + (𝒙𝒙 + 𝒚𝒚)

𝟑𝟑(𝒙𝒙 + 𝒚𝒚).

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7•3 Lesson 4


Allow students to read the problem and address the task individually. Share student responses as a class.

Example 3

A new miniature golf and arcade opened up in town. For convenient ordering, a play package is available to purchase. It includes two rounds of golf and 𝟐𝟐𝟏𝟏 arcade tokens, plus $𝟑𝟑.𝟏𝟏𝟏𝟏 off the regular price. There is a group of six friends purchasing this package. Let 𝒈𝒈 represent the cost of a round of golf, and let 𝒕𝒕 represent the cost of a token. Write two different expressions that represent the total amount this group spent. Explain how each expression describes the situation in a different way.

What two equivalent expressions could be used to represent the situation?

6(2𝑔𝑔 + 20𝑡𝑡 − 3)

Each person will pay for two rounds of golf and 20 tokens and will be discounted $3.00. This expression is six times the quantity of each friend’s cost.

12𝑔𝑔 + 120𝑡𝑡 − 18

The total cost is equal to 12 games of golf plus 120 tokens, minus $18.00 off the entire bill.


What does it mean to take the opposite of a number?

You can determine the additive inverse of a number or a multiplicative inverse. What is the opposite of 2?

−2 What is (−1)(2)?

−2 What is (−1)(𝑛𝑛)?

−𝑛𝑛 What are two mathematical expressions that represent the opposite of (2𝑎𝑎 + 3𝑏𝑏)?

(−1)(2𝑎𝑎 + 3𝑏𝑏) or −(2𝑎𝑎 + 3𝑏𝑏)

Use the distributive property to write (−1)(2𝑎𝑎 + 3𝑏𝑏) as an equivalent expression.

−2𝑎𝑎 − 3𝑏𝑏 or −2𝑎𝑎 + (−3𝑏𝑏)

To go from −2𝑎𝑎 − 3𝑏𝑏 to −(2𝑎𝑎 + 3𝑏𝑏), what process occurs?

The terms −2𝑎𝑎 and −3𝑏𝑏 are written as (−1)(2𝑎𝑎) and (−1)(3𝑏𝑏), and the −1 is factored out of their sum.

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7•3 Lesson 4


a. What is the opposite of (−𝟔𝟔𝟔𝟔+ 𝟏𝟏)?

−(−𝟔𝟔𝟔𝟔+ 𝟏𝟏)

b. Using the distributive property, write an equivalent expression for part (a).

𝟔𝟔𝟔𝟔 − 𝟏𝟏


With the class, rewrite 5𝑎𝑎 − (𝑎𝑎 − 3𝑏𝑏) applying the rules for subtracting and Example 2.

Example 5

Rewrite 𝟓𝟓𝒂𝒂 − (𝒂𝒂 − 𝟑𝟑𝒃𝒃) in standard form. Justify each step, applying the rules for subtracting and the distributive property.

𝟓𝟓𝒂𝒂 + �−(𝒂𝒂+ −𝟑𝟑𝒃𝒃)� Subtraction as adding the inverse

𝟓𝟓𝒂𝒂 + (−𝟏𝟏)(𝒂𝒂 +−𝟑𝟑𝒃𝒃) The opposite of a number is the same as multiplying by –𝟏𝟏.

𝟓𝟓𝒂𝒂 + (−𝟏𝟏)(𝒂𝒂) + (−𝟏𝟏)(−𝟑𝟑𝒃𝒃) Distributive property

𝟓𝟓𝒂𝒂 + −𝒂𝒂 + 𝟑𝟑𝒃𝒃 Multiplying by −𝟏𝟏 is the same as the opposite of the number.

𝟒𝟒𝒂𝒂 + 𝟑𝟑𝒃𝒃 Collect like terms


Encourage students to work with partners to expand each expression and collect like terms while applying the rules of subtracting and the distributive property.

Exercise 5

Expand each expression and collect like terms.

a. −𝟑𝟑(𝟐𝟐𝒑𝒑− 𝟑𝟑𝒒𝒒)

−𝟑𝟑�𝟐𝟐𝒑𝒑 + (−𝟑𝟑𝒒𝒒)� Subtraction as adding the inverse

−𝟑𝟑 ∙ 𝟐𝟐𝒑𝒑 + (−𝟑𝟑) ∙ (−𝟑𝟑𝒒𝒒) Distributive property

−𝟔𝟔𝒑𝒑 + 𝟗𝟗𝒒𝒒 Apply integer rules

b. −𝒂𝒂 − (𝒂𝒂 − 𝒃𝒃)

−𝒂𝒂 + (−(𝒂𝒂 +−𝒃𝒃)) Subtraction as adding the inverse

−𝟏𝟏𝒂𝒂 + (−𝟏𝟏(𝒂𝒂 +−𝟏𝟏𝒃𝒃)) The opposite of a number is the same as multiplying by –𝟏𝟏.

−𝟏𝟏𝒂𝒂 + (−𝟏𝟏𝒂𝒂) + 𝟏𝟏𝒃𝒃 Distributive property

−𝟐𝟐𝒂𝒂 + 𝒃𝒃 Apply integer addition rules

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7•3 Lesson 4

Closing (2 minutes)

In writing products as sums, what is happening when you take the opposite of a term or factor?

The term or factor is multiplied by −1. When using the distributive property, every term inside the parentheses is multiplied by −1.

Describe the process you used to write an expression in the form of the sum of terms as an equivalent expression in the form of a product of factors.

Writing sums as products is the opposite of writing products as sums; so, instead of distributing and multiplying, the product is being factored.


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7•3 Lesson 4

Name ___________________________________________________ Date____________________


Exit Ticket 1. Write the expression below in standard form.

3ℎ − 2(1 + 4ℎ)

2. Write the expression below as a product of two factors. 6𝑚𝑚 + 8𝑛𝑛 + 4

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7•3 Lesson 4


1. Write the expression below in standard form.

𝟑𝟑𝟑𝟑 − 𝟐𝟐(𝟏𝟏 + 𝟒𝟒𝟑𝟑)

𝟑𝟑𝟑𝟑 + (−𝟐𝟐(𝟏𝟏+ 𝟒𝟒𝟑𝟑)) Subtraction as adding the inverse

𝟑𝟑𝟑𝟑 + (−𝟐𝟐 ∙ 𝟏𝟏) + (−𝟐𝟐𝟑𝟑 ∙ 𝟒𝟒) Distributive property

𝟑𝟑𝟑𝟑 + (−𝟐𝟐) + (−𝟖𝟖𝟑𝟑) Apply integer rules

−𝟓𝟓𝟑𝟑 − 𝟐𝟐 Collect like terms

2. Write the expression below as a product of two factors.

𝟔𝟔𝒎𝒎 + 𝟖𝟖𝒏𝒏 + 𝟒𝟒

The GCF for the terms is 𝟐𝟐. Therefore, the factors are 𝟐𝟐(𝟑𝟑𝒎𝒎+ 𝟒𝟒𝒏𝒏 + 𝟐𝟐).


1. Write each expression as the product of two factors.

a. 𝟏𝟏 ∙ 𝟑𝟑 + 𝟕𝟕 ∙ 𝟑𝟑

𝟑𝟑(𝟏𝟏+ 𝟕𝟕)

b. (𝟏𝟏+ 𝟕𝟕) + (𝟏𝟏+ 𝟕𝟕) + (𝟏𝟏+ 𝟕𝟕)

𝟑𝟑(𝟏𝟏+ 𝟕𝟕)

c. 𝟐𝟐 ∙ 𝟏𝟏 + (𝟏𝟏 + 𝟕𝟕) + (𝟕𝟕 ∙ 𝟐𝟐)

𝟑𝟑(𝟏𝟏+ 𝟕𝟕)

d. 𝟑𝟑 ∙ 𝟑𝟑 + 𝟔𝟔 ∙ 𝟑𝟑

𝟑𝟑(𝟑𝟑+ 𝟔𝟔)

e. (𝟑𝟑 + 𝟔𝟔) + (𝟑𝟑 + 𝟔𝟔) + (𝟑𝟑 + 𝟔𝟔)

𝟑𝟑(𝟑𝟑+ 𝟔𝟔)

f. 𝟐𝟐𝟑𝟑 + (𝟔𝟔 + 𝟑𝟑) + 𝟔𝟔 ∙ 𝟐𝟐

𝟑𝟑(𝟑𝟑+ 𝟔𝟔)

g. 𝟒𝟒 ∙ 𝟑𝟑 + 𝟔𝟔 ∙ 𝟑𝟑

𝟑𝟑(𝟒𝟒+ 𝟔𝟔)

h. (𝟒𝟒+ 𝟔𝟔) + (𝟒𝟒+ 𝟔𝟔) + (𝐣𝐣+ 𝐤𝐤)

𝟑𝟑(𝟒𝟒+ 𝟔𝟔)

i. 𝟐𝟐𝟒𝟒 + (𝟔𝟔 + 𝟒𝟒) + 𝟐𝟐𝟔𝟔

𝟑𝟑(𝟒𝟒+ 𝟔𝟔)

2. Write each sum as a product of two factors.

a. 𝟔𝟔 ∙ 𝟕𝟕 + 𝟑𝟑 ∙ 𝟕𝟕

𝟕𝟕(𝟔𝟔+ 𝟑𝟑)

b. (𝟖𝟖+ 𝟗𝟗) + (𝟖𝟖+ 𝟗𝟗) + (𝟖𝟖+ 𝟗𝟗)

𝟑𝟑(𝟖𝟖+ 𝟗𝟗)

c. 𝟒𝟒 + (𝟏𝟏𝟐𝟐 + 𝟒𝟒) + (𝟓𝟓 ∙ 𝟒𝟒)

𝟒𝟒(𝟏𝟏+ 𝟒𝟒 + 𝟓𝟓)

d. 𝟐𝟐𝒚𝒚 ∙ 𝟑𝟑 + 𝟒𝟒 ∙ 𝟑𝟑

𝟑𝟑(𝟐𝟐𝒚𝒚+ 𝟒𝟒)

e. (𝒙𝒙 + 𝟓𝟓) + (𝒙𝒙+ 𝟓𝟓)

𝟐𝟐(𝒙𝒙 + 𝟓𝟓)

f. 𝟑𝟑𝒙𝒙 + (𝟐𝟐+ 𝒙𝒙) + 𝟓𝟓 ∙ 𝟐𝟐

𝟒𝟒(𝒙𝒙 + 𝟑𝟑)

g. 𝒇𝒇 ∙ 𝟔𝟔+ 𝒈𝒈 ∙ 𝟔𝟔

𝟔𝟔(𝒇𝒇 + 𝒈𝒈)

h. (𝒄𝒄 + 𝟏𝟏) + (𝒄𝒄 + 𝟏𝟏) + (𝒄𝒄 + 𝟏𝟏) + (𝒄𝒄 + 𝟏𝟏)

𝟒𝟒(𝒄𝒄 + 𝟏𝟏)

i. 𝟐𝟐𝒓𝒓+ 𝒓𝒓 + 𝒔𝒔 + 𝟐𝟐𝒔𝒔

𝟑𝟑(𝒓𝒓+ 𝒔𝒔)

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7•3 Lesson 4

3. Use the following rectangular array to answer the questions below.

a. Fill in the missing information.

b. Write the sum represented in the rectangular array.

𝟏𝟏𝟓𝟓𝒇𝒇 + 𝟓𝟓𝒈𝒈+ 𝟒𝟒𝟓𝟓

c. Use the missing information from part (a) to write the sum from part (b) as a product of two factors.

𝟓𝟓(𝟑𝟑𝒇𝒇 + 𝒈𝒈 + 𝟗𝟗)

4. Write the sum as a product of two factors.

a. 𝟖𝟖𝟏𝟏𝒘𝒘 + 𝟒𝟒𝟖𝟖

𝟑𝟑(𝟐𝟐𝟕𝟕𝒘𝒘+ 𝟏𝟏𝟔𝟔)

b. 𝟏𝟏𝟏𝟏 − 𝟐𝟐𝟓𝟓𝒕𝒕

𝟓𝟓(𝟐𝟐 − 𝟓𝟓𝒕𝒕)

c. 𝟏𝟏𝟐𝟐𝒂𝒂 + 𝟏𝟏𝟔𝟔𝒃𝒃 + 𝟖𝟖

𝟒𝟒(𝟑𝟑𝒂𝒂+ 𝟒𝟒𝒃𝒃 + 𝟐𝟐)

5. Xander goes to the movies with his family. Each family member buys a ticket and two boxes of popcorn. If there are five members of his family, let 𝒕𝒕 represent the cost of a ticket and 𝒑𝒑 represent the cost of a box of popcorn. Write two different expressions that represent the total amount his family spent. Explain how each expression describes the situation in a different way.

𝟓𝟓(𝒕𝒕+ 𝟐𝟐𝒑𝒑)

Five people each buy a ticket and two boxes of popcorn, so the cost is five times the quantity of a ticket and two boxes of popcorn.

𝟓𝟓𝒕𝒕 + 𝟏𝟏𝟏𝟏𝒑𝒑

There are five tickets and 𝟏𝟏𝟏𝟏 boxes of popcorn total. The total cost will be five times the cost of the tickets, plus 𝟏𝟏𝟏𝟏 times the cost of the popcorn.

𝟑𝟑𝒇𝒇 𝒈𝒈 𝟗𝟗

𝟓𝟓

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7•3 Lesson 4

6. Write each expression in standard form.

a. −𝟑𝟑(𝟏𝟏 − 𝟖𝟖𝒎𝒎− 𝟐𝟐𝒏𝒏)

−𝟑𝟑�𝟏𝟏+ (−𝟖𝟖𝒎𝒎) + (−𝟐𝟐𝒏𝒏)�

−𝟑𝟑+ 𝟐𝟐𝟒𝟒𝒎𝒎+ 𝟔𝟔𝒏𝒏

b. 𝟓𝟓 − 𝟕𝟕(−𝟒𝟒𝒒𝒒+ 𝟓𝟓)

𝟓𝟓 +−𝟕𝟕(−𝟒𝟒𝒒𝒒+ 𝟓𝟓)

𝟓𝟓 + 𝟐𝟐𝟖𝟖𝒒𝒒 + (−𝟑𝟑𝟓𝟓)

𝟐𝟐𝟖𝟖𝒒𝒒 − 𝟑𝟑𝟓𝟓 + 𝟓𝟓

𝟐𝟐𝟖𝟖𝒒𝒒 − 𝟑𝟑𝟏𝟏

c. −(𝟐𝟐𝟑𝟑− 𝟗𝟗) − 𝟒𝟒𝟑𝟑

−�𝟐𝟐𝟑𝟑+ (−𝟗𝟗)�+ (−𝟒𝟒𝟑𝟑)

−𝟐𝟐𝟑𝟑+ 𝟗𝟗 + (−𝟒𝟒𝟑𝟑)

−𝟔𝟔𝟑𝟑+ 𝟗𝟗

d. 𝟔𝟔(−𝟓𝟓𝒓𝒓 − 𝟒𝟒) − 𝟐𝟐(𝒓𝒓 − 𝟕𝟕𝒔𝒔 − 𝟑𝟑)

𝟔𝟔(−𝟓𝟓𝒓𝒓+−𝟒𝟒) + −𝟐𝟐(𝒓𝒓 − 𝟕𝟕𝒔𝒔 +−𝟑𝟑)

−𝟑𝟑𝟏𝟏𝒓𝒓+ −𝟐𝟐𝟒𝟒+ −𝟐𝟐𝒓𝒓+ 𝟏𝟏𝟒𝟒𝒔𝒔 + 𝟔𝟔

−𝟑𝟑𝟏𝟏𝒓𝒓+ −𝟐𝟐𝒓𝒓+ 𝟏𝟏𝟒𝟒𝒔𝒔 +−𝟐𝟐𝟒𝟒 + 𝟔𝟔

−𝟑𝟑𝟐𝟐𝒓𝒓+ 𝟏𝟏𝟒𝟒𝒔𝒔 − 𝟏𝟏𝟖𝟖

7. Combine like terms to write each expression in standard form.

a. (𝒓𝒓 − 𝒔𝒔) + (𝒔𝒔 − 𝒓𝒓)

𝟏𝟏

b. (−𝒓𝒓+ 𝒔𝒔) + (𝒔𝒔 − 𝒓𝒓)

−𝟐𝟐𝒓𝒓+ 𝟐𝟐𝒔𝒔

c. (−𝒓𝒓 − 𝒔𝒔)− (−𝒔𝒔 − 𝒓𝒓)

𝟏𝟏

d. (𝒓𝒓 − 𝒔𝒔) + (𝒔𝒔 − 𝒕𝒕) + (𝒕𝒕 − 𝒓𝒓)

𝟏𝟏

e. (𝒓𝒓 − 𝒔𝒔) − (𝒔𝒔 − 𝒕𝒕) − (𝒕𝒕 − 𝒓𝒓)

𝟐𝟐𝒓𝒓 − 𝟐𝟐𝒔𝒔

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Lesson 5: Using the Identity and Inverse to Write Equivalent Expressions

7•3 Lesson 5

Lesson 5: Using the Identity and Inverse to Write

Equivalent Expressions

Student Outcomes

Students recognize the identity properties of 0 and 1 and the existence of inverses (opposites and reciprocals) to write equivalent expressions.

Classwork


Students work independently to rewrite numerical expressions recalling the definitions of opposites and reciprocals.

Opening Exercise

a. In the morning, Harrison checked the temperature outside to find that it was −𝟏𝟏𝟏𝟏°𝐅𝐅. Later in the afternoon, the temperature rose 𝟏𝟏𝟏𝟏°𝐅𝐅. Write an expression representing the temperature change. What was the afternoon temperature?

−𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏; the afternoon temperature was 𝟎𝟎°𝐅𝐅.

b. Rewrite subtraction as adding the inverse for the following problems and find the sum.

i. 𝟏𝟏 − 𝟏𝟏

𝟏𝟏 + (−𝟏𝟏) = 𝟎𝟎

ii. −𝟒𝟒 − (−𝟒𝟒)

(−𝟒𝟒) + 𝟒𝟒 = 𝟎𝟎

iii. The difference of 𝟓𝟓 and 𝟓𝟓

𝟓𝟓 − 𝟓𝟓 = 𝟓𝟓 + (−𝟓𝟓) = 𝟎𝟎

iv. 𝒈𝒈 − 𝒈𝒈

𝒈𝒈 + (−𝒈𝒈) = 𝟎𝟎

c. What pattern do you notice in part (a) and (b)?

The sum of a number and its additive inverse is equal to zero.

d. Add or subtract.

i. 𝟏𝟏𝟏𝟏 + 𝟎𝟎

𝟏𝟏𝟏𝟏

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ii. 𝟎𝟎 − 𝟕𝟕

𝟎𝟎 + (−𝟕𝟕) = −𝟕𝟕

iii. −𝟒𝟒 + 𝟎𝟎

−𝟒𝟒

iv. 𝟎𝟎 + 𝒅𝒅

𝒅𝒅

v. What pattern do you notice in parts (i) through (iv)?

The sum of any quantity and zero is equal to the value of the quantity.

e. Your younger sibling runs up to you and excitedly exclaims, “I’m thinking of a number. If I add it to the number 𝟏𝟏 ten times, that is, 𝟏𝟏+ my number + my number + my number, and so on, then the answer is 𝟏𝟏. What is my number?” You almost immediately answer, “zero,” but are you sure? Can you find a different number (other than zero) that has the same property? If not, can you justify that your answer is the only correct answer?

No, there is no other number. On a number line, 𝟏𝟏 can be represented as a directed line segment that starts at 𝟎𝟎, ends at 𝟏𝟏, and has length 𝟏𝟏. Adding any other (positive or negative) number 𝒗𝒗 to 𝟏𝟏 is equivalent to attaching another directed line segment with length |𝒗𝒗| to the end of the first line segment for 𝟏𝟏:

If 𝒗𝒗 is any number other than 𝟎𝟎, then the directed line segment that represents 𝒗𝒗 will have to have some length, so 𝟏𝟏 + 𝒗𝒗 will have to be a different number on the number line. Adding 𝒗𝒗 again just takes the new sum further away from the point 𝟏𝟏 on the number line.

Discussion (5 minutes)

Discuss the following questions and conclude the Opening Exercise with definitions of opposite, additive inverse, and the identity property of zero.

In Problem 1, what is the pair of numbers called?

Opposites or additive inverses

What is the sum of a number and its opposite? It is always equal to 0.

In Problem 5, what is so special about 0?

Zero is the only number that when added with another number, the result is that other number.

This property makes zero special among all the numbers. Mathematicians have a special name for zero, called the additive identity; they call the property the additive identity property of zero.

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7•3 Lesson 5


As a class, write the sum, and then write an equivalent expression by collecting like terms and removing parentheses when possible. State the reasoning for each step.

Example 1

Write the sum, and then write an equivalent expression by collecting like terms and removing parentheses.

a. 𝟏𝟏𝒙𝒙 and −𝟏𝟏𝒙𝒙 + 𝟑𝟑

𝟏𝟏𝒙𝒙 + (−𝟏𝟏𝒙𝒙+ 𝟑𝟑)

�𝟏𝟏𝒙𝒙 + (−𝟏𝟏𝒙𝒙)�+ 𝟑𝟑 Associative property, collect like terms

𝟎𝟎 + 𝟑𝟑 Additive inverse

𝟑𝟑 Additive identity property of zero

b. 𝟏𝟏𝒙𝒙 − 𝟕𝟕 and the opposite of 𝟏𝟏𝒙𝒙

𝟏𝟏𝒙𝒙 + (−𝟕𝟕) + (−𝟏𝟏𝒙𝒙)

𝟏𝟏𝒙𝒙 + (−𝟏𝟏𝒙𝒙) + (−𝟕𝟕) Commutative property, associative property

𝟎𝟎 + (−𝟕𝟕) Additive inverse

−𝟕𝟕 Additive identity property of zero

c. The opposite of (𝟓𝟓𝒙𝒙 − 𝟏𝟏) and 𝟓𝟓𝒙𝒙

−(𝟓𝟓𝒙𝒙− 𝟏𝟏) + 𝟓𝟓𝒙𝒙

−𝟏𝟏(𝟓𝟓𝒙𝒙− 𝟏𝟏) + 𝟓𝟓𝒙𝒙 Taking the opposite is equivalent to multiplying by –𝟏𝟏.

−𝟓𝟓𝒙𝒙 + 𝟏𝟏 + 𝟓𝟓𝒙𝒙 Distributive property

(−𝟓𝟓𝒙𝒙+ 𝟓𝟓𝒙𝒙) + 𝟏𝟏 Commutative property, any order property

𝟎𝟎 + 𝟏𝟏 Additive inverse

𝟏𝟏 Additive identity property of zero


In pairs, students will take turns dictating how to write the sums while partners write what is being dictated. Students should discuss any discrepancies and explain their reasoning. Dialogue is encouraged.

Exercise 1

With a partner, take turns alternating roles as writer and speaker. The speaker verbalizes how to rewrite the sum and properties that justify each step as the writer writes what is being spoken without any input. At the end of each problem, discuss in pairs the resulting equivalent expressions.

Write the sum, and then write an equivalent expression by collecting like terms and removing parentheses whenever possible.

a. −𝟒𝟒 and 𝟒𝟒𝒃𝒃 + 𝟒𝟒

−𝟒𝟒+ (𝟒𝟒𝒃𝒃 + 𝟒𝟒)

(−𝟒𝟒+ 𝟒𝟒) + 𝟒𝟒𝒃𝒃 Any order, any grouping

𝟎𝟎 + 𝟒𝟒𝒃𝒃 Additive inverse

𝟒𝟒𝒃𝒃 Additive identity property of zero

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7•3 Lesson 5

b. 𝟑𝟑𝒙𝒙 and 𝟏𝟏 − 𝟑𝟑𝒙𝒙

𝟑𝟑𝒙𝒙 + (𝟏𝟏 − 𝟑𝟑𝒙𝒙)

𝟑𝟑𝒙𝒙 + (𝟏𝟏+ (−𝟑𝟑𝒙𝒙)) Subtraction as adding the inverse

�𝟑𝟑𝒙𝒙 + (−𝟑𝟑𝒙𝒙)�+ 𝟏𝟏 Any order, any grouping

𝟎𝟎 + 𝟏𝟏 Additive inverse

𝟏𝟏 Additive identity property of zero

c. The opposite of 𝟒𝟒𝒙𝒙 and −𝟓𝟓 + 𝟒𝟒𝒙𝒙

−𝟒𝟒𝒙𝒙 + (−𝟓𝟓 + 𝟒𝟒𝒙𝒙)

(−𝟒𝟒𝒙𝒙+ 𝟒𝟒𝒙𝒙) + (−𝟓𝟓) Any order, any grouping

𝟎𝟎 + (−𝟓𝟓) Additive inverse

−𝟓𝟓 Additive identity property of zero

d. The opposite of −𝟏𝟏𝟎𝟎𝒕𝒕 and 𝒕𝒕 − 𝟏𝟏𝟎𝟎𝒕𝒕

𝟏𝟏𝟎𝟎𝒕𝒕+ (𝒕𝒕 − 𝟏𝟏𝟎𝟎𝒕𝒕)

�𝟏𝟏𝟎𝟎𝒕𝒕+ (−𝟏𝟏𝟎𝟎𝒕𝒕)�+ 𝒕𝒕 Any order, any grouping

𝟎𝟎 + 𝒕𝒕 Additive inverse

𝒕𝒕 Additive identity property of zero

e. The opposite of (−𝟕𝟕− 𝟒𝟒𝒗𝒗) and −𝟒𝟒𝒗𝒗

−(−𝟕𝟕− 𝟒𝟒𝒗𝒗) + (−𝟒𝟒𝒗𝒗)

−𝟏𝟏(−𝟕𝟕− 𝟒𝟒𝒗𝒗) + (−𝟒𝟒𝒗𝒗) Taking the opposite is equivalent to multiplying by –𝟏𝟏.

𝟕𝟕 + 𝟒𝟒𝒗𝒗 + (−𝟒𝟒𝒗𝒗) Distributive property

𝟕𝟕 + 𝟎𝟎 Any grouping, additive inverse

𝟕𝟕 Additive identity property of zero


Students should complete the first five problems independently and then discuss as a class.

Example 2

�𝟑𝟑𝟒𝟒� × �𝟒𝟒𝟑𝟑� = 𝟏𝟏

𝟒𝟒 × 𝟏𝟏𝟒𝟒 = 𝟏𝟏

𝟏𝟏𝟗𝟗

× 𝟗𝟗 = 𝟏𝟏

�−𝟏𝟏𝟑𝟑�× −𝟑𝟑 = 𝟏𝟏

�−𝟏𝟏𝟓𝟓�× �−𝟓𝟓

𝟏𝟏� = 𝟏𝟏

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7•3 Lesson 5

What are these pairs of numbers called?

Reciprocals

What is another term for reciprocal?

The multiplicative inverse

What happens to the sign of the expression when converting it to its multiplicative inverse?

There is no change to the sign. For example, the multiplicative inverse of −2 is �− 12�. The negative

sign remains the same.

What can you conclude from the pattern in the answers?

The product of a number and its multiplicative inverse is equal to 1. Earlier, we saw that 0 is a special number because it is the only number that when added to another number,

results in that number again. Can you explain why the number 1 is also special?

One is the only number that when multiplied with another number, results in that number again.

This property makes 1 special among all the numbers. Mathematicians have a special name for 1, the multiplicative identity; they call the property the multiplicative identity property of one.

As an extension, ask students if there are any other special numbers that they have learned. Students should respond: Yes; −1 has the property that multiplying a number by it is the same as taking the opposite of the number. Share with students that they are going to learn later in this module about another special number called pi.

As a class, write the product, and then write an equivalent expression in standard form. State the properties for each step. After discussing questions, review the properties and definitions in the Lesson Summary emphasizing the multiplicative identity property of one and the multiplicative inverse.

Write the product, and then write the expression in standard form by removing parentheses and combining like terms. Justify each step.

a. The multiplicative inverse of 𝟏𝟏𝟓𝟓

and �𝟏𝟏𝒙𝒙 − 𝟏𝟏𝟓𝟓�

𝟓𝟓�𝟏𝟏𝒙𝒙 − 𝟏𝟏𝟓𝟓�

𝟓𝟓(𝟏𝟏𝒙𝒙)− 𝟓𝟓 ⋅ 𝟏𝟏𝟓𝟓 Distributive property

𝟏𝟏𝟎𝟎𝒙𝒙 − 𝟏𝟏 Multiplicative inverses

b. The multiplicative inverse of 𝟏𝟏 and (𝟏𝟏𝒙𝒙 + 𝟒𝟒)

�𝟏𝟏𝟏𝟏� (𝟏𝟏𝒙𝒙+ 𝟒𝟒)

�𝟏𝟏𝟏𝟏� (𝟏𝟏𝒙𝒙) + �𝟏𝟏𝟏𝟏� (𝟒𝟒) Distributive property

𝟏𝟏𝒙𝒙 + 𝟏𝟏 Multiplicative inverses, multiplication

𝒙𝒙 + 𝟏𝟏 Multiplicative identity property of one

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7•3 Lesson 5

c. The multiplicative inverse of � 𝟏𝟏𝟑𝟑𝒙𝒙+𝟓𝟓� and

𝟏𝟏𝟑𝟑

(𝟑𝟑𝒙𝒙 + 𝟓𝟓) ⋅ 𝟏𝟏𝟑𝟑

𝟑𝟑𝒙𝒙�𝟏𝟏𝟑𝟑� + 𝟓𝟓�𝟏𝟏𝟑𝟑� Distributive property

𝟏𝟏𝒙𝒙 + 𝟓𝟓𝟑𝟑 Multiplicative inverse

𝒙𝒙 + 𝟓𝟓𝟑𝟑 Multiplicative identity property of one


As in Exercise 1, have students work in pairs to rewrite the expressions, taking turns being the speaker and writer.

Exercise 2

Write the product, and then write the expression in standard form by removing parentheses and combining like terms. Justify each step.

a. The reciprocal of 𝟑𝟑 and −𝟏𝟏𝟔𝟔 − 𝟑𝟑𝒙𝒙

�𝟏𝟏𝟑𝟑� �−𝟏𝟏𝟔𝟔+ (−𝟑𝟑𝒙𝒙)� Rewrite subtraction as an addition problem

�𝟏𝟏𝟑𝟑� (−𝟏𝟏𝟔𝟔) + �𝟏𝟏𝟑𝟑� (−𝟑𝟑𝒙𝒙) Distributive property

−𝟏𝟏𝟔𝟔 − 𝟏𝟏𝒙𝒙 Multiplicative inverse

−𝟏𝟏𝟔𝟔 − 𝒙𝒙 Multiplicative identity property of one

b. The multiplicative inverse of 𝟒𝟒 and 𝟒𝟒𝟒𝟒 − 𝟏𝟏𝟎𝟎

�𝟏𝟏𝟒𝟒� �𝟒𝟒𝟒𝟒+ (−𝟏𝟏𝟎𝟎)� Rewrite subtraction as an addition problem

�𝟏𝟏𝟒𝟒� (𝟒𝟒𝟒𝟒) + �𝟏𝟏𝟒𝟒� (−𝟏𝟏𝟎𝟎) Distributive property

𝟏𝟏𝟒𝟒 + (−𝟓𝟓) Multiplicative inverse

𝟒𝟒 − 𝟓𝟓 Multiplicative identity property of one

c. The multiplicative inverse of −𝟏𝟏𝟏𝟏 and 𝟏𝟏 − 𝟏𝟏

𝟏𝟏 𝒋𝒋

(−𝟏𝟏)�𝟏𝟏 + �−𝟏𝟏𝟏𝟏 𝒋𝒋�� Rewrite subtraction as an addition problem

(−𝟏𝟏)(𝟏𝟏) + (−𝟏𝟏) �−𝟏𝟏𝟏𝟏 𝒋𝒋� Distributive property

−𝟏𝟏𝟏𝟏+ 𝟏𝟏𝒋𝒋 Multiplicative inverse

−𝟏𝟏𝟏𝟏+ 𝒋𝒋 Multiplicative identity property of one

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7•3 Lesson 5

Closing (3 minutes)

What are the other terms for opposites and reciprocals, and what are the general rules of their sums and products? Additive inverse and multiplicative inverse; the sum of additive inverses equals 0; the product of

multiplicative inverses equals 1.

What do the additive identity property of zero and the multiplicative identity property of one state?

The additive identity property of zero states that zero is the only number that when added to another number, the result is again that number. The multiplicative identity property of one states that one is the only number that when multiplied with another number, the result is that number again.


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7•3 Lesson 5

Name ___________________________________________________ Date____________________

Lesson 5: Using the Identity and Inverse to Write Equivalent

Expressions

Exit Ticket 1. Find the sum of 5𝑥𝑥 + 20 and the opposite of 20. Write an equivalent expression in standard form. Justify each

step.

2. For 5𝑥𝑥 + 20 and the multiplicative inverse of 5, write the product and then write the expression in standard form, if possible. Justify each step.

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7•3 Lesson 5


1. Find the sum of 𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟎𝟎 and the opposite of 𝟏𝟏𝟎𝟎. Write an equivalent expression in standard form. Justify each step.

(𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟎𝟎) + (−𝟏𝟏𝟎𝟎)

𝟓𝟓𝒙𝒙 + �𝟏𝟏𝟎𝟎+ (−𝟏𝟏𝟎𝟎)� Associative property of addition

𝟓𝟓𝒙𝒙 + 𝟎𝟎 Additive inverse

𝟓𝟓𝒙𝒙 Additive identity property of zero

2. For 𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟎𝟎 and the multiplicative inverse of 𝟓𝟓, write the product and then write the expression in standard form, if possible. Justify each step.

(𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟎𝟎) �𝟏𝟏𝟓𝟓�

(𝟓𝟓𝒙𝒙) �𝟏𝟏𝟓𝟓�+ 𝟏𝟏𝟎𝟎�𝟏𝟏𝟓𝟓� Distributive property

𝟏𝟏𝒙𝒙 + 𝟒𝟒 Multiplicative inverses, multiplication

𝒙𝒙 + 𝟒𝟒 Multiplicative identity property of one


1. Fill in the missing parts.

a. The sum of 𝟏𝟏𝟔𝟔 − 𝟓𝟓 and the opposite of 𝟏𝟏𝟔𝟔

(𝟏𝟏𝟔𝟔 − 𝟓𝟓) + (−𝟏𝟏𝟔𝟔)

�𝟏𝟏𝟔𝟔 + (−𝟓𝟓)�+ (−𝟏𝟏𝟔𝟔) Rewrite subtraction as addition

𝟏𝟏𝟔𝟔 + (−𝟏𝟏𝟔𝟔) + (−𝟓𝟓) Regrouping/any order (or commutative property of addition)

𝟎𝟎 + (−𝟓𝟓) Additive inverse

−𝟓𝟓 Additive identity property of zero

b. The product of −𝟏𝟏𝟔𝟔 + 𝟏𝟏𝟒𝟒 and the multiplicative inverse of −𝟏𝟏

(−𝟏𝟏𝟔𝟔 + 𝟏𝟏𝟒𝟒) �−𝟏𝟏𝟏𝟏�

(−𝟏𝟏𝟔𝟔) �−𝟏𝟏𝟏𝟏� + (𝟏𝟏𝟒𝟒) �−𝟏𝟏

𝟏𝟏� Distributive property

𝟏𝟏𝟔𝟔 + (−𝟕𝟕) Multiplicative inverse, multiplication

𝟏𝟏𝟔𝟔 − 𝟕𝟕 Adding the additive inverse is the same as subtraction

𝟔𝟔 − 𝟕𝟕 Multiplicative identity property of one

2. Write the sum, and then rewrite the expression in standard form by removing parentheses and collecting like terms.

a. 𝟏𝟏 and 𝒑𝒑 − 𝟏𝟏

𝟏𝟏 + (𝒑𝒑 − 𝟏𝟏)

𝟏𝟏 + (−𝟏𝟏) + 𝒑𝒑

𝟎𝟎 + 𝒑𝒑

𝒑𝒑

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7•3 Lesson 5

b. 𝟏𝟏𝟎𝟎𝟏𝟏 + 𝟑𝟑 and −𝟑𝟑

(𝟏𝟏𝟎𝟎𝟏𝟏 + 𝟑𝟑) + (−𝟑𝟑)

𝟏𝟏𝟎𝟎𝟏𝟏 + �𝟑𝟑+ (−𝟑𝟑)�

𝟏𝟏𝟎𝟎𝟏𝟏 + 𝟎𝟎

𝟏𝟏𝟎𝟎𝟏𝟏

c. −𝒙𝒙 − 𝟏𝟏𝟏𝟏 and the opposite of −𝟏𝟏𝟏𝟏

�−𝒙𝒙 + (−𝟏𝟏𝟏𝟏)�+ 𝟏𝟏𝟏𝟏

−𝒙𝒙 + �(−𝟏𝟏𝟏𝟏) + (𝟏𝟏𝟏𝟏)�

−𝒙𝒙 + 𝟎𝟎

−𝒙𝒙

d. The opposite of 𝟒𝟒𝒙𝒙 and 𝟑𝟑+ 𝟒𝟒𝒙𝒙

(−𝟒𝟒𝒙𝒙) + (𝟑𝟑+ 𝟒𝟒𝒙𝒙)

�(−𝟒𝟒𝒙𝒙) + 𝟒𝟒𝒙𝒙�+ 𝟑𝟑

𝟎𝟎 + 𝟑𝟑

𝟑𝟑

e. 𝟏𝟏𝒈𝒈 and the opposite of (𝟏𝟏 − 𝟏𝟏𝒈𝒈)

𝟏𝟏𝒈𝒈 + �−(𝟏𝟏 − 𝟏𝟏𝒈𝒈)�

𝟏𝟏𝒈𝒈 + (−𝟏𝟏) + 𝟏𝟏𝒈𝒈

𝟏𝟏𝒈𝒈 + 𝟏𝟏𝒈𝒈 + (−𝟏𝟏)

𝟒𝟒𝒈𝒈 + (−𝟏𝟏)

𝟒𝟒𝒈𝒈 − 𝟏𝟏

3. Write the product, and then rewrite the expression in standard form by removing parentheses and collecting like terms.

a. 𝟕𝟕𝟒𝟒 − 𝟏𝟏 and the multiplicative inverse of 𝟕𝟕

�𝟕𝟕𝟒𝟒 + (−𝟏𝟏)� �𝟏𝟏𝟕𝟕�

�𝟏𝟏𝟕𝟕� (𝟕𝟕𝟒𝟒) + �

𝟏𝟏𝟕𝟕� (−𝟏𝟏)

𝟒𝟒 −𝟏𝟏𝟕𝟕

b. The multiplicative inverse of −𝟓𝟓 and 𝟏𝟏𝟎𝟎𝒗𝒗 − 𝟓𝟓

�−𝟏𝟏𝟓𝟓� (𝟏𝟏𝟎𝟎𝒗𝒗 − 𝟓𝟓)

�−𝟏𝟏𝟓𝟓� (𝟏𝟏𝟎𝟎𝒗𝒗) + �−

𝟏𝟏𝟓𝟓� (−𝟓𝟓)

−𝟏𝟏𝒗𝒗+ 𝟏𝟏

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7•3 Lesson 5

c. 𝟗𝟗 − 𝒃𝒃 and the multiplicative inverse of 𝟗𝟗

�𝟗𝟗+ (−𝒃𝒃)� �𝟏𝟏𝟗𝟗�

�𝟏𝟏𝟗𝟗� (𝟗𝟗) + �

𝟏𝟏𝟗𝟗� (−𝒃𝒃)

𝟏𝟏 −𝟏𝟏𝟗𝟗𝒃𝒃

d. The multiplicative inverse of 𝟏𝟏𝟒𝟒 and 𝟓𝟓𝒕𝒕 − 𝟏𝟏𝟒𝟒

𝟒𝟒�𝟓𝟓𝒕𝒕 −𝟏𝟏𝟒𝟒�

𝟒𝟒(𝟓𝟓𝒕𝒕) + 𝟒𝟒�−𝟏𝟏𝟒𝟒�

𝟏𝟏𝟎𝟎𝒕𝒕 − 𝟏𝟏

e. The multiplicative inverse of − 𝟏𝟏𝟏𝟏𝟎𝟎𝒙𝒙 and 𝟏𝟏

𝟏𝟏𝟎𝟎𝒙𝒙−

𝟏𝟏𝟏𝟏𝟎𝟎

(−𝟏𝟏𝟎𝟎𝒙𝒙)�𝟏𝟏𝟏𝟏𝟎𝟎𝒙𝒙

−𝟏𝟏𝟏𝟏𝟎𝟎�

(−𝟏𝟏𝟎𝟎𝒙𝒙)�𝟏𝟏𝟏𝟏𝟎𝟎𝒙𝒙

�+ (−𝟏𝟏𝟎𝟎𝒙𝒙) �−𝟏𝟏𝟏𝟏𝟎𝟎�

−𝟏𝟏+ 𝒙𝒙

4. Write the expressions in standard form.

a. 𝟏𝟏𝟒𝟒

(𝟒𝟒𝒙𝒙 + 𝟖𝟖)

𝟏𝟏𝟒𝟒

(𝟒𝟒𝒙𝒙) +𝟏𝟏𝟒𝟒

(𝟖𝟖)

𝒙𝒙 + 𝟏𝟏

b. 𝟏𝟏𝟏𝟏

(𝒓𝒓 − 𝟏𝟏)

𝟏𝟏𝟏𝟏

(𝒓𝒓) +𝟏𝟏𝟏𝟏

(−𝟏𝟏)

𝟏𝟏𝟏𝟏𝒓𝒓 − 𝟏𝟏

c. 𝟒𝟒𝟓𝟓

(𝒙𝒙 + 𝟏𝟏)

𝟒𝟒𝟓𝟓

(𝒙𝒙) +𝟒𝟒𝟓𝟓

(𝟏𝟏)

𝟒𝟒𝟓𝟓𝒙𝒙 +

𝟒𝟒𝟓𝟓

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7•3 Lesson 5

d. 𝟏𝟏𝟖𝟖

(𝟏𝟏𝒙𝒙 + 𝟒𝟒)

𝟏𝟏𝟖𝟖

(𝟏𝟏𝒙𝒙) +𝟏𝟏𝟖𝟖

(𝟒𝟒)

𝟏𝟏𝟒𝟒𝒙𝒙 +

𝟏𝟏𝟏𝟏

e. 𝟑𝟑𝟒𝟒

(𝟓𝟓𝒙𝒙 − 𝟏𝟏)

𝟑𝟑𝟒𝟒

(𝟓𝟓𝒙𝒙) +𝟑𝟑𝟒𝟒

(−𝟏𝟏)

𝟏𝟏𝟓𝟓𝟒𝟒𝒙𝒙 −

𝟑𝟑𝟒𝟒

f. 𝟏𝟏𝟓𝟓

(𝟏𝟏𝟎𝟎𝒙𝒙 − 𝟓𝟓) − 𝟑𝟑

𝟏𝟏𝟓𝟓

(𝟏𝟏𝟎𝟎𝒙𝒙) +𝟏𝟏𝟓𝟓

(−𝟓𝟓) + (−𝟑𝟑)

𝟏𝟏𝒙𝒙 + (−𝟏𝟏) + (−𝟑𝟑)

𝟏𝟏𝒙𝒙 − 𝟒𝟒

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Lesson 6: Collecting Rational Number Like Terms

7•3 Lesson 6


Student Outcomes

Students rewrite rational number expressions by collecting like terms and combining them by repeated use of the distributive property.

Classwork


Students work in pairs to write the expressions in standard form. Reconvene as a class and review the steps taken to rewrite the expressions, justifying each step verbally as you go.

Opening Exercise

Solve each problem, leaving your answers in standard form. Show your steps.

a. Terry weighs 𝟒𝟒𝟒𝟒 𝐤𝐤𝐤𝐤. Janice weighs 𝟐𝟐𝟑𝟑𝟒𝟒 𝐤𝐤𝐤𝐤 less than Terry. What is their combined weight?

𝟒𝟒𝟒𝟒+ �𝟒𝟒𝟒𝟒 − 𝟐𝟐𝟑𝟑𝟒𝟒� = 𝟖𝟖𝟒𝟒 − 𝟐𝟐 − 𝟑𝟑𝟒𝟒 = 𝟕𝟕𝟖𝟖 − 𝟑𝟑

𝟒𝟒 = 𝟕𝟕𝟕𝟕𝟏𝟏𝟒𝟒. Their combined weight is 𝟕𝟕𝟕𝟕𝟏𝟏𝟒𝟒 𝐤𝐤𝐤𝐤.

b. 𝟐𝟐𝟐𝟐𝟑𝟑 − 𝟏𝟏𝟏𝟏𝟐𝟐 −𝟒𝟒𝟓𝟓

𝟖𝟖𝟑𝟑−𝟑𝟑𝟐𝟐−𝟒𝟒𝟓𝟓

𝟖𝟖𝟒𝟒𝟑𝟑𝟒𝟒

−𝟒𝟒𝟓𝟓𝟑𝟑𝟒𝟒

−𝟐𝟐𝟒𝟒𝟑𝟑𝟒𝟒

𝟏𝟏𝟏𝟏𝟑𝟑𝟒𝟒

c. 𝟏𝟏𝟓𝟓

+ (−𝟒𝟒)

−𝟑𝟑𝟒𝟒𝟓𝟓

d. 𝟒𝟒�𝟑𝟑𝟓𝟓�

𝟒𝟒𝟏𝟏�𝟑𝟑𝟓𝟓�

𝟏𝟏𝟐𝟐𝟓𝟓

𝟐𝟐𝟐𝟐𝟓𝟓

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7•3 Lesson 6

e. Mr. Jackson bought 𝟏𝟏𝟑𝟑𝟓𝟓 𝐥𝐥𝐥𝐥. of beef. He cooked 𝟑𝟑𝟒𝟒 of it for lunch. How much does he have left?

If he cooked 𝟑𝟑𝟒𝟒

of it for lunch, he had 𝟏𝟏𝟒𝟒

of the original amount left. Since �𝟏𝟏𝟑𝟑𝟓𝟓��𝟏𝟏𝟒𝟒� = 𝟖𝟖

𝟓𝟓 ⋅𝟏𝟏𝟒𝟒 = 𝟐𝟐

𝟓𝟓, he had 𝟐𝟐𝟓𝟓

𝐥𝐥𝐥𝐥.

left. Teachers: You can also show your students how to write the answer as one expression.

�𝟏𝟏𝟑𝟑𝟓𝟓� �𝟏𝟏 −

𝟑𝟑𝟒𝟒�

How is the process of writing equivalent expressions by combining like terms in this Opening Exercise different from the previous lesson?

There are additional steps to find common denominators, convert mixed numbers to improper numbers (in some cases), and convert back.


Example 1

Rewrite the expression in standard form by collecting like terms.

𝟐𝟐𝟑𝟑𝒏𝒏 −

𝟑𝟑𝟒𝟒𝒏𝒏+

𝟏𝟏𝟔𝟔𝒏𝒏 + 𝟐𝟐

𝟐𝟐𝟗𝟗𝒏𝒏

𝟐𝟐𝟒𝟒𝟑𝟑𝟔𝟔

𝒏𝒏 −𝟐𝟐𝟕𝟕𝟑𝟑𝟔𝟔

𝒏𝒏 +𝟔𝟔𝟑𝟑𝟔𝟔

𝒏𝒏 + 𝟐𝟐𝟖𝟖𝟑𝟑𝟔𝟔

𝒏𝒏

𝟐𝟐𝟏𝟏𝟏𝟏𝟑𝟑𝟔𝟔

𝒏𝒏

What are various strategies for adding, subtracting, multiplying, and dividing rational numbers?

Find common denominators; change from mixed numbers and whole numbers to improper fractions, and then convert back.


Walk around as students work independently. Have students check their answers with a partner. Address any unresolved questions.

Scaffolding: Students who need a challenge could tackle the following problem:

12𝑎𝑎 + 2

23𝑏𝑏 +

15−

14𝑎𝑎 − 1

12𝑏𝑏 +

35

+34𝑎𝑎 − 4

−45𝑏𝑏

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7•3 Lesson 6

Exercise 1

For the following exercises, predict how many terms the resulting expression will have after collecting like terms. Then, write the expression in standard form by collecting like terms.

a. 𝟐𝟐𝟓𝟓𝒈𝒈 −

𝟏𝟏𝟔𝟔− 𝒈𝒈 +

𝟑𝟑𝟏𝟏𝟒𝟒

𝒈𝒈 −𝟒𝟒𝟓𝟓

There will be two terms.

𝟐𝟐𝟓𝟓𝒈𝒈 − 𝟏𝟏𝒈𝒈 +

𝟑𝟑𝟏𝟏𝟒𝟒

𝒈𝒈 −𝟏𝟏𝟔𝟔−𝟒𝟒𝟓𝟓

�𝟐𝟐𝟓𝟓− 𝟏𝟏 +

𝟑𝟑𝟏𝟏𝟒𝟒�𝒈𝒈 − �

𝟏𝟏𝟔𝟔

+𝟒𝟒𝟓𝟓�

−𝟑𝟑𝟏𝟏𝟒𝟒

𝒈𝒈 −𝟐𝟐𝟗𝟗𝟑𝟑𝟒𝟒

b. 𝒊𝒊 + 𝟔𝟔𝒊𝒊 − 𝟑𝟑𝟕𝟕 𝒊𝒊 + 𝟏𝟏

𝟑𝟑𝒉𝒉 + 𝟏𝟏𝟐𝟐 𝒊𝒊 − 𝒉𝒉 + 𝟏𝟏

𝟒𝟒𝒉𝒉

There will be two terms.

𝟏𝟏𝟑𝟑𝒉𝒉 +

𝟏𝟏𝟒𝟒𝒉𝒉 − 𝒉𝒉 + 𝒊𝒊 −

𝟑𝟑𝟕𝟕𝒊𝒊 + 𝟔𝟔𝒊𝒊 +

𝟏𝟏𝟐𝟐𝒊𝒊

�𝟏𝟏𝟑𝟑

+𝟏𝟏𝟒𝟒

+ (−𝟏𝟏)�𝒉𝒉 + �𝟏𝟏 −𝟑𝟑𝟕𝟕

+ 𝟔𝟔+𝟏𝟏𝟐𝟐� 𝒊𝒊

−𝟓𝟓𝟏𝟏𝟐𝟐

𝒉𝒉 + 𝟕𝟕𝟏𝟏𝟏𝟏𝟒𝟒

𝒊𝒊


Read the problem as a class and give students time to set up their own expressions. Reconvene as a class to address each expression.

Example 2

At a store, a shirt was marked down in price by $𝟏𝟏𝟒𝟒.𝟒𝟒𝟒𝟒. A pair of pants doubled in price. Following these changes, the price of every item in the store was cut in half. Write two different expressions that represent the new cost of the items, using 𝒔𝒔 for the cost of each shirt and 𝒑𝒑 for the cost of a pair of pants. Explain the different information each one shows.

For the cost of a shirt:

𝟏𝟏𝟐𝟐

(𝒔𝒔 − 𝟏𝟏𝟒𝟒) The cost of each shirt is 𝟏𝟏𝟐𝟐 of the quantity of the original cost of the shirt, minus 𝟏𝟏𝟒𝟒.

𝟏𝟏𝟐𝟐𝒔𝒔 − 𝟓𝟓 The cost of each shirt is half off the original price, minus 𝟓𝟓, since half of 𝟏𝟏𝟒𝟒 is 𝟓𝟓.

For the cost of a pair of pants:

𝟏𝟏𝟐𝟐

(𝟐𝟐𝒑𝒑) The cost of each pair of pants is half off double the price.

𝒑𝒑 The cost of each pair of pants is the original cost because 𝟏𝟏𝟐𝟐

is the multiplicative inverse of 𝟐𝟐.

Describe a situation in which either of the two expressions in each case would be more useful.

Answers may vary. For example, 𝑝𝑝 would be more useful than 12

(2𝑝𝑝) because it is converted back to an

isolated variable, in this case the original cost.

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7•3 Lesson 6


Write two different expressions that represent the total cost of the items if tax was 𝟏𝟏𝟏𝟏𝟒𝟒

of the original price. Explain the

different information each shows.

For the cost of a shirt:

𝟏𝟏𝟐𝟐

(𝒔𝒔 − 𝟏𝟏𝟒𝟒) +𝟏𝟏𝟏𝟏𝟒𝟒

𝒔𝒔 The cost of each shirt is 𝟏𝟏𝟐𝟐

of the quantity of the original cost of the shirt, minus 𝟏𝟏𝟒𝟒, plus 𝟏𝟏𝟏𝟏𝟒𝟒

of the cost of the shirt.

𝟑𝟑𝟓𝟓𝒔𝒔 − 𝟓𝟓 The cost of each shirt is 𝟑𝟑

𝟓𝟓 of the original price (because it is 𝟏𝟏𝟐𝟐𝒔𝒔 + 𝟏𝟏

𝟏𝟏𝟒𝟒𝒔𝒔 = 𝟔𝟔𝟏𝟏𝟒𝟒𝒔𝒔), minus 𝟓𝟓, since half of 𝟏𝟏𝟒𝟒

is 𝟓𝟓.

For the cost of a pair of pants:

𝟏𝟏𝟐𝟐

(𝟐𝟐𝒑𝒑) +𝟏𝟏𝟏𝟏𝟒𝟒

𝒑𝒑 The cost of each pair of pants is half off double the price plus 𝟏𝟏𝟏𝟏𝟒𝟒

of the cost of a pair of pants.

𝟏𝟏 𝟏𝟏𝟏𝟏𝟒𝟒𝒑𝒑 The cost of each pair of pants is 𝟏𝟏 𝟏𝟏

𝟏𝟏𝟒𝟒 (because 𝟏𝟏𝒑𝒑 + 𝟏𝟏𝟏𝟏𝟒𝟒𝒑𝒑 = 𝟏𝟏 𝟏𝟏

𝟏𝟏𝟒𝟒𝒑𝒑) times the number of pair of pants.


Example 3

Write this expression in standard form by collecting like terms. Justify each step.

𝟓𝟓𝟏𝟏𝟑𝟑− �𝟑𝟑

𝟏𝟏𝟑𝟑� �𝟏𝟏𝟐𝟐𝒙𝒙 −

𝟏𝟏𝟒𝟒�

𝟏𝟏𝟔𝟔𝟑𝟑

+ �− 𝟏𝟏𝟒𝟒𝟑𝟑� �𝟏𝟏

𝟐𝟐𝒙𝒙� + �− 𝟏𝟏𝟒𝟒

𝟑𝟑� �− 𝟏𝟏

𝟒𝟒� Write mixed numbers as improper fractions, then distribute.

𝟏𝟏𝟔𝟔𝟑𝟑

+ �− 𝟓𝟓𝟑𝟑𝒙𝒙� +

𝟓𝟓𝟔𝟔

Any grouping (associative) and arithmetic rules for multiplying rational numbers

−𝟓𝟓𝟑𝟑𝒙𝒙 + �𝟑𝟑𝟐𝟐𝟔𝟔 + 𝟓𝟓

𝟔𝟔� Commutative property and associative property of addition, collect like terms

−𝟓𝟓𝟑𝟑𝒙𝒙 + 𝟑𝟑𝟕𝟕

𝟔𝟔 Apply arithmetic rule for adding rational numbers

A student says he created an equivalent expression by first finding this difference: 5 13 − 3 1

3 . Is he correct? Why or why not? Although they do appear to be like terms, taking the difference would be incorrect. In the

expression �3 13� �

12 𝑥𝑥 −

14�, 3 1

3 must be distributed before applying any other operation in this problem.

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7•3 Lesson 6

How should 3 13 be written before being distributed?

The mixed number can be rewritten as an improper fraction 103

. It is not necessary to convert the mixed

number, but it makes the process more efficient and increases the likelihood of getting a correct answer.


Walk around as students work independently. Have students check their answers with a partner. Address any unresolved questions.

Exercise 3

Rewrite the following expressions in standard form by finding the product and collecting like terms.

a. −𝟔𝟔𝟏𝟏𝟑𝟑 −𝟏𝟏𝟐𝟐 �

𝟏𝟏𝟐𝟐 + 𝒚𝒚�

−𝟔𝟔𝟏𝟏𝟑𝟑

+ �−𝟏𝟏𝟐𝟐� �𝟏𝟏𝟐𝟐�+ �−

𝟏𝟏𝟐𝟐�𝒚𝒚

−𝟔𝟔𝟏𝟏𝟑𝟑

+ �−𝟏𝟏𝟒𝟒�+ �−

𝟏𝟏𝟐𝟐𝒚𝒚�

−𝟏𝟏𝟐𝟐𝒚𝒚 − �𝟔𝟔

𝟏𝟏𝟑𝟑

+𝟏𝟏𝟒𝟒�

−𝟏𝟏𝟐𝟐𝒚𝒚 − �𝟔𝟔

𝟒𝟒𝟏𝟏𝟐𝟐

+𝟑𝟑𝟏𝟏𝟐𝟐�

−𝟏𝟏𝟐𝟐𝒚𝒚 − 𝟔𝟔

𝟕𝟕𝟏𝟏𝟐𝟐

b. 𝟐𝟐𝟑𝟑

+𝟏𝟏𝟑𝟑�𝟏𝟏𝟒𝟒𝒇𝒇 − 𝟏𝟏

𝟏𝟏𝟑𝟑�

𝟐𝟐𝟑𝟑

+𝟏𝟏𝟑𝟑�𝟏𝟏𝟒𝟒𝒇𝒇�+

𝟏𝟏𝟑𝟑�−

𝟒𝟒𝟑𝟑�

𝟐𝟐𝟑𝟑

+𝟏𝟏𝟏𝟏𝟐𝟐

𝒇𝒇 −𝟒𝟒𝟗𝟗

𝟏𝟏𝟏𝟏𝟐𝟐

𝒇𝒇 + �𝟔𝟔𝟗𝟗−𝟒𝟒𝟗𝟗�

𝟏𝟏𝟏𝟏𝟐𝟐

𝒇𝒇 +𝟐𝟐𝟗𝟗


Example 4

Model how to write the expression in standard form using rules of rational numbers.

𝒙𝒙𝟐𝟐𝟒𝟒

+𝟐𝟐𝒙𝒙𝟓𝟓

+𝒙𝒙 + 𝟏𝟏𝟐𝟐

+𝟑𝟑𝒙𝒙 − 𝟏𝟏𝟏𝟏𝟒𝟒

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7•3 Lesson 6

What are other equivalent expressions of 𝑥𝑥20

? How do you know?

Other expressions include 1𝑥𝑥20

and 120

𝑥𝑥 because of the arithmetic rules of rational numbers.

What about 120𝑥𝑥

? How do you know?

It is not equivalent because if 𝑥𝑥 = 2, the value of the expression is 140

, which does not equal 110

.

How can the distributive property be used in this problem?

For example, it can be used to factor out 120

from each term of the expression.

Or, for example, it can be used to distribute 110

: 3𝑥𝑥−110

=110

(3𝑥𝑥 − 1) =3𝑥𝑥10

−110

.

Below are two solutions. Explore both with the class.

𝒙𝒙𝟐𝟐𝟒𝟒

+𝟒𝟒(𝟐𝟐𝒙𝒙)𝟐𝟐𝟒𝟒

+𝟏𝟏𝟒𝟒(𝒙𝒙+ 𝟏𝟏)

𝟐𝟐𝟒𝟒+𝟐𝟐(𝟑𝟑𝒙𝒙 − 𝟏𝟏)

𝟐𝟐𝟒𝟒

𝒙𝒙 + 𝟖𝟖𝒙𝒙 + 𝟏𝟏𝟒𝟒𝒙𝒙 + 𝟏𝟏𝟒𝟒 + 𝟔𝟔𝒙𝒙 − 𝟐𝟐𝟐𝟐𝟒𝟒

𝟐𝟐𝟓𝟓𝒙𝒙 + 𝟖𝟖𝟐𝟐𝟒𝟒

𝟓𝟓𝟒𝟒𝒙𝒙 +

𝟐𝟐𝟓𝟓

𝟏𝟏𝟐𝟐𝟒𝟒

𝒙𝒙 +𝟐𝟐𝟓𝟓𝒙𝒙 +

𝟏𝟏𝟐𝟐𝒙𝒙 +

𝟏𝟏𝟐𝟐

+𝟑𝟑𝟏𝟏𝟒𝟒

𝒙𝒙 −𝟏𝟏𝟏𝟏𝟒𝟒

�𝟏𝟏𝟐𝟐𝟒𝟒

+𝟐𝟐𝟓𝟓

+𝟏𝟏𝟐𝟐

+𝟑𝟑𝟏𝟏𝟒𝟒�𝒙𝒙 + �

𝟏𝟏𝟐𝟐−

𝟏𝟏𝟏𝟏𝟒𝟒�

�𝟏𝟏𝟐𝟐𝟒𝟒

+𝟖𝟖𝟐𝟐𝟒𝟒

+𝟏𝟏𝟒𝟒𝟐𝟐𝟒𝟒

+𝟔𝟔𝟐𝟐𝟒𝟒�𝒙𝒙 + �

𝟓𝟓𝟏𝟏𝟒𝟒

−𝟏𝟏𝟏𝟏𝟒𝟒�

𝟓𝟓𝟒𝟒𝒙𝒙 +

𝟐𝟐𝟓𝟓

Ask students to evaluate the original expression and the answers when 𝑥𝑥 = 20 to see if they get the same number.

Evaluate the original expression and the answers when 𝒙𝒙 = 𝟐𝟐𝟒𝟒. Do you get the same number?

𝒙𝒙𝟐𝟐𝟒𝟒

+𝟐𝟐𝒙𝒙𝟓𝟓

+𝒙𝒙 + 𝟏𝟏𝟐𝟐

+𝟑𝟑𝒙𝒙 − 𝟏𝟏𝟏𝟏𝟒𝟒

𝟐𝟐𝟒𝟒𝟐𝟐𝟒𝟒

+𝟐𝟐(𝟐𝟐𝟒𝟒)𝟓𝟓

+𝟐𝟐𝟒𝟒+ 𝟏𝟏𝟐𝟐

+𝟑𝟑(𝟐𝟐𝟒𝟒) − 𝟏𝟏

𝟏𝟏𝟒𝟒

𝟏𝟏 + 𝟖𝟖+𝟐𝟐𝟏𝟏𝟐𝟐

+𝟓𝟓𝟗𝟗𝟏𝟏𝟒𝟒

𝟗𝟗 +𝟏𝟏𝟒𝟒𝟓𝟓𝟏𝟏𝟒𝟒

+𝟓𝟓𝟗𝟗𝟏𝟏𝟒𝟒

𝟗𝟗+𝟏𝟏𝟔𝟔𝟒𝟒𝟏𝟏𝟒𝟒

𝟗𝟗 + 𝟏𝟏𝟔𝟔𝟒𝟒𝟏𝟏𝟒𝟒

𝟐𝟐𝟓𝟓𝟐𝟐𝟓𝟓

𝟓𝟓𝟒𝟒𝒙𝒙 +

𝟐𝟐𝟓𝟓

𝟓𝟓𝟒𝟒

(𝟐𝟐𝟒𝟒) +𝟐𝟐𝟓𝟓

𝟐𝟐𝟓𝟓+𝟐𝟐𝟓𝟓

𝟐𝟐𝟓𝟓𝟐𝟐𝟓𝟓

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7•3 Lesson 6

Important: After students evaluate both expressions for 𝑥𝑥 = 20, ask them which expression was easier. (The expression in standard form is easier.) Explain to students: When you are asked on a standardized test to simplify an expression, you must put the expression in standard form because standard form is often much simpler to evaluate and read. This curriculum is specific and will often tell you the form (such as standard form) it wants you to write the expression in for an answer.


Allow students to work independently.

Exercise 4

Rewrite the following expression in standard form by finding common denominators and collecting like terms.

𝟐𝟐𝒉𝒉𝟑𝟑−𝒉𝒉𝟗𝟗

+𝒉𝒉 − 𝟒𝟒𝟔𝟔

𝟔𝟔(𝟐𝟐𝒉𝒉)𝟏𝟏𝟖𝟖

−𝟐𝟐(𝒉𝒉)𝟏𝟏𝟖𝟖

+𝟑𝟑(𝒉𝒉 − 𝟒𝟒)

𝟏𝟏𝟖𝟖

𝟏𝟏𝟐𝟐𝒉𝒉 − 𝟐𝟐𝒉𝒉 + 𝟑𝟑𝒉𝒉 − 𝟏𝟏𝟐𝟐𝟏𝟏𝟖𝟖

𝟏𝟏𝟑𝟑𝒉𝒉 − 𝟏𝟏𝟐𝟐𝟏𝟏𝟖𝟖

𝟏𝟏𝟑𝟑𝟏𝟏𝟖𝟖

𝒉𝒉 −𝟐𝟐𝟑𝟑

Example 5 (Optional, 5 minutes)

Give students a minute to observe the expression and decide how to begin rewriting it in standard form.

Example 5

Rewrite the following expression in standard form.

𝟐𝟐(𝟑𝟑𝒙𝒙 − 𝟒𝟒)𝟔𝟔

−𝟓𝟓𝒙𝒙 + 𝟐𝟐𝟖𝟖

How can we start to rewrite this problem?

There are various ways to start rewriting this expression, including using the distributive property,

renaming 26 , rewriting the subtraction as an addition, distributing the negative in the second term,

rewriting each term as a fraction (e.g., 26

(3𝑥𝑥 − 4) − �5𝑥𝑥8

+28�), or finding the lowest common

denominator.

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7•3 Lesson 6

Method 1: Method 2a: Method 2b: Method 3:

𝟏𝟏(𝟑𝟑𝒙𝒙− 𝟒𝟒)𝟑𝟑

−𝟓𝟓𝒙𝒙 + 𝟐𝟐𝟖𝟖

𝟖𝟖(𝟑𝟑𝒙𝒙 − 𝟒𝟒)𝟐𝟐𝟒𝟒

−𝟑𝟑(𝟓𝟓𝒙𝒙 + 𝟐𝟐)

𝟐𝟐𝟒𝟒

�(𝟐𝟐𝟒𝟒𝒙𝒙 − 𝟑𝟑𝟐𝟐) − (𝟏𝟏𝟓𝟓𝒙𝒙 + 𝟔𝟔)�𝟐𝟐𝟒𝟒

(𝟐𝟐𝟒𝟒𝒙𝒙− 𝟑𝟑𝟐𝟐 − 𝟏𝟏𝟓𝟓𝒙𝒙 − 𝟔𝟔)𝟐𝟐𝟒𝟒

𝟗𝟗𝒙𝒙 − 𝟑𝟑𝟖𝟖𝟐𝟐𝟒𝟒

𝟗𝟗𝒙𝒙𝟐𝟐𝟒𝟒

−𝟑𝟑𝟖𝟖𝟐𝟐𝟒𝟒

𝟑𝟑𝟖𝟖𝒙𝒙 −

𝟏𝟏𝟗𝟗𝟏𝟏𝟐𝟐

𝟔𝟔𝒙𝒙 − 𝟖𝟖𝟔𝟔

−𝟓𝟓𝒙𝒙 + 𝟐𝟐𝟖𝟖

𝟒𝟒(𝟔𝟔𝒙𝒙 − 𝟖𝟖)𝟐𝟐𝟒𝟒

−𝟑𝟑(𝟓𝟓𝒙𝒙+ 𝟐𝟐)

𝟖𝟖

(𝟐𝟐𝟒𝟒𝒙𝒙 − 𝟑𝟑𝟐𝟐 − 𝟏𝟏𝟓𝟓𝒙𝒙 − 𝟔𝟔)𝟐𝟐𝟒𝟒

𝟗𝟗𝒙𝒙 − 𝟑𝟑𝟖𝟖𝟐𝟐𝟒𝟒

𝟗𝟗𝒙𝒙𝟐𝟐𝟒𝟒

−𝟑𝟑𝟖𝟖𝟐𝟐𝟒𝟒

𝟑𝟑𝟖𝟖𝒙𝒙 −

𝟏𝟏𝟗𝟗𝟏𝟏𝟐𝟐

𝟔𝟔𝟔𝟔𝒙𝒙 −

𝟖𝟖𝟔𝟔−𝟓𝟓𝟖𝟖𝒙𝒙 −

𝟐𝟐𝟖𝟖

𝒙𝒙 −𝟒𝟒𝟑𝟑−𝟓𝟓𝒙𝒙𝟖𝟖−𝟏𝟏𝟒𝟒

𝟏𝟏𝒙𝒙 −𝟓𝟓𝟖𝟖𝒙𝒙 −

𝟒𝟒𝟑𝟑−𝟏𝟏𝟒𝟒

𝟑𝟑𝟖𝟖𝒙𝒙 −

𝟏𝟏𝟔𝟔𝟏𝟏𝟐𝟐

−𝟑𝟑𝟏𝟏𝟐𝟐

𝟑𝟑𝟖𝟖𝒙𝒙 −

𝟏𝟏𝟗𝟗𝟏𝟏𝟐𝟐

𝟏𝟏𝟑𝟑

(𝟑𝟑𝒙𝒙− 𝟒𝟒) − �𝟓𝟓𝒙𝒙𝟖𝟖

+𝟏𝟏𝟒𝟒�

𝒙𝒙 −𝟒𝟒𝟑𝟑−𝟓𝟓𝟖𝟖𝒙𝒙 −

𝟏𝟏𝟒𝟒

𝟏𝟏𝒙𝒙 −𝟓𝟓𝟖𝟖𝒙𝒙 −

𝟒𝟒𝟑𝟑−𝟏𝟏𝟒𝟒

𝟑𝟑𝟖𝟖𝒙𝒙 −

𝟏𝟏𝟔𝟔𝟏𝟏𝟐𝟐

−𝟑𝟑𝟏𝟏𝟐𝟐

𝟑𝟑𝟖𝟖𝒙𝒙 −

𝟏𝟏𝟗𝟗𝟏𝟏𝟐𝟐

Which method(s) keep(s) the numbers in the expression in integer form? Why would this be important to note?

Finding the lowest common denominator would keep the number in integer form; this is important because working with the terms would be more convenient.

Is one method better than the rest of the methods?

No, it is by preference; however, the properties of addition and multiplication must be used properly.

Are these expressions equivalent: 38𝑥𝑥, 3𝑥𝑥8

, and 38𝑥𝑥

? How do you know?

The first two expressions are equivalent, but the third one, 38𝑥𝑥

, is not. If you substitute a value other

than zero or one (such as 𝑥𝑥 = 2), the values of the first expressions are the same, 23

. The value of the

third expression is 316

.

What are some common errors that could occur when rewriting this expression in standard form? Some common errors may include distributing only to one term in the parentheses, forgetting to

multiply the negative sign to all the terms in the parentheses, incorrectly reducing fractions, and/or adjusting the common denominator but not the numerator.

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7•3 Lesson 6

Exercise 5 (Optional, 3 minutes)

Allow students to work independently. Have students share the various ways they started to rewrite the problem.

Exercise 5

Write the following expression in standard form.

𝟐𝟐𝒙𝒙 − 𝟏𝟏𝟏𝟏𝟒𝟒

−𝟑𝟑(𝒙𝒙 − 𝟐𝟐)

𝟏𝟏𝟒𝟒

𝟓𝟓(𝟐𝟐𝒙𝒙 − 𝟏𝟏𝟏𝟏)𝟐𝟐𝟒𝟒

−𝟐𝟐 ∙ 𝟑𝟑(𝒙𝒙 − 𝟐𝟐)

𝟐𝟐𝟒𝟒

(𝟏𝟏𝟒𝟒𝒙𝒙 − 𝟓𝟓𝟓𝟓) − 𝟔𝟔(𝒙𝒙 − 𝟐𝟐)𝟐𝟐𝟒𝟒

𝟏𝟏𝟒𝟒𝒙𝒙 − 𝟓𝟓𝟓𝟓 − 𝟔𝟔𝒙𝒙 + 𝟏𝟏𝟐𝟐𝟐𝟐𝟒𝟒

𝟒𝟒𝒙𝒙 − 𝟒𝟒𝟑𝟑𝟐𝟐𝟒𝟒

𝟏𝟏𝟓𝟓𝒙𝒙 − 𝟐𝟐

𝟑𝟑𝟐𝟐𝟒𝟒

Closing (2 minutes)

Jane says combining like terms is much harder to do when the coefficients and constant terms are not integers. Why do you think Jane feels this way?

There are usually more steps, including finding common denominators, converting mixed numbers to improper fractions, etc.


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7•3 Lesson 6

Name ___________________________________________________ Date____________________


Exit Ticket

For the problem 15𝑔𝑔 −

110

− 𝑔𝑔 + 1310

𝑔𝑔 −110

, Tyson created an equivalent expression using the following steps.

15𝑔𝑔 + −1𝑔𝑔 + 1

310

𝑔𝑔 + −1

10+ −

110

−45𝑔𝑔 + 1

110

Is his final expression equivalent to the initial expression? Show how you know. If the two expressions are not equivalent, find Tyson’s mistake and correct it.

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7•3 Lesson 6


For the problem 𝟏𝟏𝟓𝟓𝒈𝒈 −

𝟏𝟏𝟏𝟏𝟒𝟒

− 𝒈𝒈 + 𝟏𝟏𝟑𝟑𝟏𝟏𝟒𝟒

𝒈𝒈 −𝟏𝟏𝟏𝟏𝟒𝟒

, Tyson created an equivalent expression using the following steps.

𝟏𝟏𝟓𝟓𝒈𝒈+ −𝟏𝟏𝒈𝒈 + 𝟏𝟏

𝟑𝟑𝟏𝟏𝟒𝟒

𝒈𝒈 + −𝟏𝟏𝟏𝟏𝟒𝟒

+ −𝟏𝟏𝟏𝟏𝟒𝟒

−𝟒𝟒𝟓𝟓𝒈𝒈 + 𝟏𝟏

𝟏𝟏𝟏𝟏𝟒𝟒

Is his final expression equivalent to the initial expression? Show how you know. If the two expressions are not equivalent, find Tyson’s mistake and correct it.

No, he added the first two terms correctly, but he forgot the third term and added to the other like terms.

If 𝒈𝒈 = 𝟏𝟏𝟒𝟒,

𝟏𝟏𝟓𝟓𝒈𝒈 +−𝟏𝟏𝒈𝒈 + 𝟏𝟏

𝟑𝟑𝟏𝟏𝟒𝟒

𝒈𝒈+ −𝟏𝟏𝟏𝟏𝟒𝟒

+−𝟏𝟏𝟏𝟏𝟒𝟒

𝟏𝟏𝟓𝟓

(𝟏𝟏𝟒𝟒) +−𝟏𝟏(𝟏𝟏𝟒𝟒) + 𝟏𝟏𝟑𝟑𝟏𝟏𝟒𝟒

(𝟏𝟏𝟒𝟒) + −𝟏𝟏𝟏𝟏𝟒𝟒

+−𝟏𝟏𝟏𝟏𝟒𝟒

𝟐𝟐+ (−𝟏𝟏𝟒𝟒) + 𝟏𝟏𝟑𝟑 + �−𝟐𝟐𝟏𝟏𝟒𝟒�

𝟒𝟒𝟒𝟒𝟓𝟓

−𝟒𝟒𝟓𝟓𝒈𝒈 + 𝟏𝟏

𝟏𝟏𝟏𝟏𝟒𝟒

−𝟒𝟒𝟓𝟓

(𝟏𝟏𝟒𝟒) + 𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒

−𝟖𝟖 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒

−𝟔𝟔𝟗𝟗𝟏𝟏𝟒𝟒

The expressions are not equal.

He should factor out the 𝒈𝒈 and place parentheses around the values using the distributive property in order to make it obvious which rational numbers need to be combined.

𝟏𝟏𝟓𝟓𝒈𝒈+ −𝟏𝟏𝒈𝒈 + 𝟏𝟏

𝟑𝟑𝟏𝟏𝟒𝟒

𝒈𝒈 + −𝟏𝟏𝟏𝟏𝟒𝟒

+ −𝟏𝟏𝟏𝟏𝟒𝟒

�𝟏𝟏𝟓𝟓𝒈𝒈 +−𝟏𝟏𝒈𝒈 + 𝟏𝟏

𝟑𝟑𝟏𝟏𝟒𝟒

𝒈𝒈� + �−𝟏𝟏𝟏𝟏𝟒𝟒

+ −𝟏𝟏𝟏𝟏𝟒𝟒�

�𝟏𝟏𝟓𝟓

+ −𝟏𝟏+ 𝟏𝟏𝟑𝟑𝟏𝟏𝟒𝟒�𝒈𝒈+ �−

𝟐𝟐𝟏𝟏𝟒𝟒�

�𝟐𝟐𝟏𝟏𝟒𝟒

+𝟑𝟑𝟏𝟏𝟒𝟒�𝒈𝒈 + �−

𝟏𝟏𝟓𝟓�

𝟏𝟏𝟐𝟐𝒈𝒈 −

𝟏𝟏𝟓𝟓


1. Write the indicated expressions.

a. 𝟏𝟏𝟐𝟐𝒎𝒎 inches in feet

𝟏𝟏𝟐𝟐𝒎𝒎 ×

𝟏𝟏𝟏𝟏𝟐𝟐

=𝟏𝟏𝟐𝟐𝟒𝟒

𝒎𝒎. It is 𝟏𝟏𝟐𝟐𝟒𝟒

𝒎𝒎 𝐟𝐟𝐟𝐟.

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7•3 Lesson 6

b. The perimeter of a square with 𝟐𝟐𝟑𝟑𝒈𝒈 𝐜𝐜𝐜𝐜 sides

𝟒𝟒 × 𝟐𝟐𝟑𝟑𝒈𝒈 = 𝟖𝟖

𝟑𝟑𝒈𝒈. The perimeter is 𝟖𝟖𝟑𝟑𝒈𝒈 𝐜𝐜𝐜𝐜.

c. The number of pounds in 𝟗𝟗 𝐨𝐨𝐨𝐨.

𝟗𝟗 × 𝟏𝟏𝟏𝟏𝟔𝟔 = 𝟗𝟗

𝟏𝟏𝟔𝟔. It is 𝟗𝟗𝟏𝟏𝟔𝟔

𝐥𝐥𝐥𝐥.

d. The average speed of a train that travels 𝒙𝒙 miles in 𝟑𝟑𝟒𝟒 hour

𝑹𝑹 = 𝑫𝑫𝑻𝑻;

𝒙𝒙𝟑𝟑𝟒𝟒

=𝟒𝟒𝟑𝟑𝒙𝒙. The average speed of the train is

𝟒𝟒𝟑𝟑𝒙𝒙 miles per hour.

e. Devin is 𝟏𝟏𝟏𝟏𝟒𝟒 years younger than Eli. April is 𝟏𝟏𝟓𝟓

as old as Devin. Jill is 𝟓𝟓 years older than April. If Eli is 𝑬𝑬 years

old, what is Jill’s age in terms of 𝑬𝑬?

𝑫𝑫 = 𝑬𝑬− 𝟏𝟏𝟏𝟏𝟒𝟒, 𝑨𝑨 = 𝑫𝑫𝟓𝟓, 𝑨𝑨 + 𝟓𝟓 = 𝑱𝑱, so 𝑱𝑱 = �𝑫𝑫𝟓𝟓� + 𝟓𝟓. 𝑱𝑱 = 𝟏𝟏

𝟓𝟓 (𝑬𝑬 − 𝟏𝟏𝟏𝟏𝟒𝟒) + 𝟓𝟓. 𝑱𝑱 = 𝑬𝑬𝟓𝟓 + 𝟒𝟒𝟑𝟑𝟒𝟒.

2. Rewrite the expressions by collecting like terms.

a. 𝟏𝟏𝟐𝟐𝒌𝒌 −

𝟑𝟑𝟖𝟖𝒌𝒌

𝟒𝟒𝟖𝟖𝒌𝒌 −

𝟑𝟑𝟖𝟖𝒌𝒌

𝟏𝟏𝟖𝟖𝒌𝒌

b. 𝟐𝟐𝟐𝟐𝟓𝟓

+𝟕𝟕𝟐𝟐𝟏𝟏𝟓𝟓

𝟔𝟔𝟐𝟐𝟏𝟏𝟓𝟓

+𝟕𝟕𝟐𝟐𝟏𝟏𝟓𝟓

𝟏𝟏𝟑𝟑𝟐𝟐𝟏𝟏𝟓𝟓

c. −𝟏𝟏𝟑𝟑𝒂𝒂 −

𝟏𝟏𝟐𝟐𝒃𝒃 −

𝟑𝟑𝟒𝟒 + 𝟏𝟏

𝟐𝟐𝒃𝒃 −𝟐𝟐𝟑𝟑𝒃𝒃 + 𝟓𝟓

𝟔𝟔𝒂𝒂

−𝟏𝟏𝟑𝟑𝒂𝒂 +

𝟓𝟓𝟔𝟔𝒂𝒂 −

𝟏𝟏𝟐𝟐𝒃𝒃 +

𝟏𝟏𝟐𝟐𝒃𝒃 −

𝟐𝟐𝟑𝟑𝒃𝒃 −

𝟑𝟑𝟒𝟒

−𝟐𝟐𝟔𝟔𝒂𝒂 +

𝟓𝟓𝟔𝟔𝒂𝒂 −

𝟐𝟐𝟑𝟑𝒃𝒃 −

𝟑𝟑𝟒𝟒

𝟏𝟏𝟐𝟐𝒂𝒂 −

𝟐𝟐𝟑𝟑𝒃𝒃 −

𝟑𝟑𝟒𝟒

d. −𝒑𝒑 + 𝟑𝟑𝟓𝟓𝒒𝒒 −

𝟏𝟏𝟏𝟏𝟒𝟒𝒒𝒒 + 𝟏𝟏

𝟗𝟗 −𝟏𝟏𝟗𝟗𝒑𝒑 + 𝟐𝟐𝟏𝟏𝟑𝟑𝒑𝒑

−𝒑𝒑 −𝟏𝟏𝟗𝟗𝒑𝒑 + 𝟐𝟐

𝟏𝟏𝟑𝟑𝒑𝒑 +

𝟑𝟑𝟓𝟓𝒒𝒒 −

𝟏𝟏𝟏𝟏𝟒𝟒

𝒒𝒒 +𝟏𝟏𝟗𝟗

−𝟗𝟗𝟗𝟗𝒑𝒑 −

𝟏𝟏𝟗𝟗𝒑𝒑 + 𝟐𝟐

𝟑𝟑𝟗𝟗𝒑𝒑 +

𝟔𝟔𝟏𝟏𝟒𝟒

𝒒𝒒 −𝟏𝟏𝟏𝟏𝟒𝟒

𝒒𝒒 +𝟏𝟏𝟗𝟗

𝟏𝟏𝟏𝟏𝟗𝟗𝒑𝒑 +

𝟓𝟓𝟏𝟏𝟒𝟒

𝒒𝒒 +𝟏𝟏𝟗𝟗

𝟏𝟏𝟐𝟐𝟗𝟗𝒑𝒑 +

𝟏𝟏𝟐𝟐𝒒𝒒 +

𝟏𝟏𝟗𝟗

e. 𝟓𝟓𝟕𝟕𝒚𝒚 −

𝒚𝒚𝟏𝟏𝟒𝟒

𝟏𝟏𝟒𝟒𝟏𝟏𝟒𝟒

𝒚𝒚 −𝟏𝟏𝟏𝟏𝟒𝟒

𝒚𝒚

𝟗𝟗𝟏𝟏𝟒𝟒

𝒚𝒚

f. 𝟑𝟑𝒏𝒏𝟖𝟖−𝒏𝒏𝟒𝟒

+ 𝟐𝟐𝒏𝒏𝟐𝟐

𝟑𝟑𝒏𝒏𝟖𝟖−𝟐𝟐𝒏𝒏𝟖𝟖

+ 𝟐𝟐𝟒𝟒𝒏𝒏𝟖𝟖

𝟐𝟐𝟓𝟓𝒏𝒏𝟖𝟖

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7•3 Lesson 6

3. Rewrite the expressions by using the distributive property and collecting like terms.

a. 𝟒𝟒𝟓𝟓

(𝟏𝟏𝟓𝟓𝒙𝒙 − 𝟓𝟓)

𝟏𝟏𝟐𝟐𝒙𝒙 − 𝟒𝟒

b. 𝟒𝟒𝟓𝟓�𝟏𝟏𝟒𝟒𝒄𝒄 − 𝟓𝟓�

𝟏𝟏𝟓𝟓𝒄𝒄 − 𝟒𝟒

c. 𝟐𝟐𝟒𝟒𝟓𝟓𝒗𝒗 −𝟐𝟐𝟑𝟑�𝟒𝟒𝒗𝒗 + 𝟏𝟏𝟏𝟏𝟔𝟔�

𝟐𝟐𝟏𝟏𝟓𝟓

𝒗𝒗 −𝟕𝟕𝟗𝟗

d. 𝟖𝟖 − 𝟒𝟒�𝟏𝟏𝟖𝟖𝟐𝟐 − 𝟑𝟑𝟏𝟏𝟐𝟐�

−𝟏𝟏𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐

e. 𝟏𝟏𝟕𝟕

(𝟏𝟏𝟒𝟒𝒙𝒙 + 𝟕𝟕) − 𝟓𝟓

𝟐𝟐𝒙𝒙 − 𝟒𝟒

f. 𝟏𝟏𝟓𝟓

(𝟓𝟓𝒙𝒙 − 𝟏𝟏𝟓𝟓) − 𝟐𝟐𝒙𝒙

−𝒙𝒙 − 𝟑𝟑

g. 𝟏𝟏𝟒𝟒

(𝒑𝒑 + 𝟒𝟒) +𝟑𝟑𝟓𝟓

(𝒑𝒑 − 𝟏𝟏)

𝟏𝟏𝟕𝟕𝟐𝟐𝟒𝟒

𝒑𝒑 +𝟐𝟐𝟓𝟓

h. 𝟕𝟕𝟖𝟖

(𝒘𝒘 + 𝟏𝟏) +𝟓𝟓𝟔𝟔

(𝒘𝒘 − 𝟑𝟑)

𝟒𝟒𝟏𝟏𝟐𝟐𝟒𝟒

𝒘𝒘−𝟑𝟑𝟗𝟗𝟐𝟐𝟒𝟒

or 𝟒𝟒𝟏𝟏𝟐𝟐𝟒𝟒

𝒘𝒘−𝟏𝟏𝟑𝟑𝟖𝟖

i. 𝟒𝟒𝟓𝟓

(𝒄𝒄 − 𝟏𝟏) − 𝟏𝟏𝟖𝟖

(𝟐𝟐𝒄𝒄 + 𝟏𝟏)

𝟏𝟏𝟏𝟏𝟐𝟐𝟒𝟒

𝒄𝒄 −𝟑𝟑𝟕𝟕𝟒𝟒𝟒𝟒

j. 𝟐𝟐𝟑𝟑�𝒉𝒉 +

𝟑𝟑𝟒𝟒� − 𝟏𝟏

𝟑𝟑�𝒉𝒉 +

𝟑𝟑𝟒𝟒�

𝟏𝟏𝟑𝟑𝒉𝒉 +

𝟏𝟏𝟒𝟒

k. 𝟐𝟐𝟑𝟑�𝒉𝒉 +

𝟑𝟑𝟒𝟒� − 𝟐𝟐

𝟑𝟑�𝒉𝒉 − 𝟑𝟑

𝟒𝟒�

𝟏𝟏

l. 𝟐𝟐𝟑𝟑�𝒉𝒉 +

𝟑𝟑𝟒𝟒� +

𝟐𝟐𝟑𝟑�𝒉𝒉 − 𝟑𝟑

𝟒𝟒�

𝟒𝟒𝟑𝟑𝒉𝒉

m. 𝒌𝒌𝟐𝟐−𝟒𝟒𝒌𝒌𝟓𝟓− 𝟑𝟑

−𝟑𝟑𝒌𝒌𝟏𝟏𝟒𝟒

− 𝟑𝟑

n. 𝟑𝟑𝟑𝟑 + 𝟐𝟐𝟕𝟕

+𝟑𝟑 − 𝟒𝟒𝟏𝟏𝟒𝟒

𝟏𝟏𝟐𝟐𝟑𝟑

o. 𝟗𝟗𝒙𝒙 − 𝟒𝟒𝟏𝟏𝟒𝟒

+𝟑𝟑𝒙𝒙 + 𝟐𝟐𝟓𝟓

𝟑𝟑𝒙𝒙𝟐𝟐

or 𝟏𝟏𝟏𝟏𝟐𝟐𝒙𝒙

p. 𝟑𝟑(𝟓𝟓𝒈𝒈 − 𝟏𝟏)

𝟒𝟒−𝟐𝟐𝒈𝒈 + 𝟕𝟕𝟔𝟔

𝟑𝟑𝟓𝟓𝟏𝟏𝟐𝟐

𝒈𝒈 − 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐

q. −𝟑𝟑𝟑𝟑 + 𝟏𝟏𝟓𝟓 + 𝟑𝟑 − 𝟓𝟓

𝟐𝟐 + 𝟕𝟕𝟏𝟏𝟒𝟒

−𝟑𝟑𝟏𝟏𝟒𝟒

− 𝟐𝟐

r. 𝟗𝟗𝒘𝒘𝟔𝟔

+𝟐𝟐𝒘𝒘 − 𝟕𝟕

𝟑𝟑−𝒘𝒘 − 𝟓𝟓𝟒𝟒

𝟐𝟐𝟑𝟑𝒘𝒘− 𝟏𝟏𝟑𝟑𝟏𝟏𝟐𝟐

𝟐𝟐𝟑𝟑𝟏𝟏𝟐𝟐

𝒘𝒘−𝟏𝟏𝟑𝟑𝟏𝟏𝟐𝟐

s. 𝟏𝟏 + 𝒇𝒇𝟓𝟓

−𝟏𝟏 + 𝒇𝒇𝟑𝟑

+𝟑𝟑 − 𝒇𝒇𝟔𝟔

𝟏𝟏𝟏𝟏𝟑𝟑𝟒𝟒

−𝟑𝟑𝟏𝟏𝟒𝟒

𝒇𝒇

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Topic B: Solve Problems Using Expressions, Equations, and Inequalities

7 G R A D E

Mathematics Curriculum Topic B

Solve Problems Using Expressions, Equations, and Inequalities

7.EE.B.3, 7.EE.B.4, 7.G.B.5

Focus Standards: 7.EE.B.3

Solve multi-step real-life and mathematical problems posed with positive and negative rational numbers in any form (whole numbers, fractions, and decimals), using tools strategically. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate; and assess the reasonableness of answers using mental computation and estimation strategies. For example: If a woman making $25 an hour gets a 10% raise, she will make an additional 1/10 of her salary an hour, or $2.50, for a new salary of $27.50. If you want to place a towel bar 9 3/4 inches long in the center of a door that is 27 1/2 inches wide, you will need to place the bar about 9 inches from each edge; this estimate can be used as a check on the exact computation.

7.EE.B.4 Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. a. Solve word problems leading to equations of the form 𝑝𝑝𝑝𝑝 + 𝑞𝑞 = 𝑟𝑟 and

𝑝𝑝(𝑝𝑝 + 𝑞𝑞) = 𝑟𝑟, where 𝑝𝑝, 𝑞𝑞, and 𝑟𝑟 are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach. For example, the perimeter of a rectangle is 54 cm. Its length is 6 cm. What is its width?

b. Solve word problems leading to inequalities of the form 𝑝𝑝𝑝𝑝 + 𝑞𝑞 > 𝑟𝑟 and 𝑝𝑝𝑝𝑝 + 𝑞𝑞 < 𝑟𝑟, where 𝑝𝑝, 𝑞𝑞, and 𝑟𝑟 are specific rational numbers. Graph the solution set of the inequality and interpret it in the context of the problem. For example: As a salesperson, you are paid $50 per week plus $3 per sale. This week you want your pay to be at least $100. Write an inequality for the number of sales you need to make and describe the solutions.

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7•3 Topic B

Topic B: Solve Problems Using Expressions, Equations, and Inequalities

7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and use them to solve simple equations for an unknown angle in a figure.


Lesson 7: Understanding Equations (P)1

Lessons 8–9: Using If-Then Moves in Solving Equations (P)

Lessons 10–11: Angle Problems and Solving Equations (P)

Lesson 12: Properties of Inequalities (E)

Lesson 13: Inequalities (P)

Lesson 14: Solving Inequalities (P)

Lesson 15: Graphing Solutions to Inequalities (P)

Topic B begins in Lesson 7 with students evaluating equations and problems modeled with equations for given rational number values to determine whether the value makes a true or false number sentence. In Lessons 8 and 9, students are given problems of perimeter; total cost; age comparisons; and distance, rate, and time to solve. Students will discover that modeling these types of problems with an equation becomes an efficient approach to solving the problem, especially when the problem contains rational numbers (7.EE.B.3, 7.EE.B.4a). Students apply the properties of equality to isolate the variable in these equations as well as those created to model missing angle problems in Lessons 10 and 11. All problems provide a real-world or mathematical context so that students can connect the (abstract) variable, or letter, to the number that it actually represents in the problem. The number already exists; students just need to find it.

Lesson 12 introduces students to situations that are modeled in the form 𝑝𝑝𝑝𝑝 + 𝑞𝑞 > 𝑟𝑟 and 𝑝𝑝𝑝𝑝 + 𝑞𝑞 < 𝑟𝑟. Initially, students start by translating from verbal to algebraic, choosing the inequality symbol that best represents the given situation. Students then find the number(s) that make each inequality true. To better understand how to solve an inequality containing a variable, students look at statements comparing numbers in Lesson 13. They discover when (and why) multiplying by a negative number reverses the inequality symbol when this symbol is preserved. In Lesson 14, students extend the idea of isolating the variable in an equation to solve problems modeled with inequalities using the properties of inequality. This topic concludes with students modeling inequality solutions on a number line and interpreting what each solution means within the context of the problem (7.EE.B.4b).


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7•3 Lesson 7

Lesson 7: Understanding Equations


Student Outcomes

Students understand that an equation is a statement of equality between two expressions. Students build an algebraic expression using the context of a word problem and use that expression to write

an equation that can be used to solve the word problem.

Lesson Notes Students are asked to substitute a number for the variable to check whether it is a solution to the equation.

This lesson focuses on students building an equation that can be used to solve a word problem. The variable (letter) in an equation is a placeholder for a number. The equations might be true for some numbers and false for others. A solution to an equation is a number that makes the equation true. The emphasis of this lesson is for students to build an algebraic expression and set it equal to a number to form an equation that can be used to solve a word problem. As part of the activity, students are asked to check whether a number (or set of numbers) is a solution to the equation. Solving an equation algebraically is left for future lessons.

The definitions presented below form the foundation of the next few lessons in this topic. Please review these carefully to help you understand the structure of the lessons.

EQUATION: An equation is a statement of equality between two expressions.

If 𝐴𝐴 and 𝐵𝐵 are two expressions in the variable 𝑥𝑥, then 𝐴𝐴 = 𝐵𝐵 is an equation in the variable 𝑥𝑥.

Students sometimes have trouble keeping track of what is an expression and what is an equation. An expression never includes an equal sign (=) and can be thought of as part of a sentence. The expression 3 + 4 read aloud is, “Three plus four,” which is only a phrase in a possible sentence. Equations, on the other hand, always have an equal sign, which is a symbol for the verb is. The equation 3 + 4 = 7 read aloud is, “Three plus four is seven,” which expresses a complete thought (i.e., a sentence).

Number sentences—equations with numbers only—are special among all equations.

NUMBER SENTENCE: A number sentence is a statement of equality (or inequality) between two numerical expressions.

A number sentence is by far the most concrete version of an equation. It also has the very important property that it is always true or always false, and it is this property that distinguishes it from a generic equation. Examples include 3 +4 = 7 (true) and 3 + 3 = 7 (false). This important property guarantees the ability to check whether or not a number is a solution to an equation with a variable: just substitute a number into the variable. The resulting number sentence is either true or it is false. If the number sentence is true, the number is a solution to the equation. For that reason, number sentences are the first and most important type of equation that students need to understand.1

Of course, we are mostly interested in numbers that make equations into true number sentences, and we have a special name for such numbers: a solution.

1Note that sentence is a legitimate mathematical term, not just a student-friendly version of equation meant for Grade 1 or Grade 2. To see what the term ultimately becomes, take a look at http://mathworld.wolfram.com/Sentence.html.

7

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http://mathworld.wolfram.com/Sentence.html

7•3 Lesson 7


SOLUTION: A solution to an equation with one variable is a number that, when substituted for all instances of the variable in both expressions, makes the equation a true number sentence.

Classwork

Opening Exercise (10 minutes) Opening Exercise

Your brother is going away to college, so you no longer have to share a bedroom. You decide to redecorate a wall by hanging two new posters on the wall. The wall is 𝟏𝟏𝟏𝟏 feet wide, and each poster is four feet wide. You want to place the posters on the wall so that the distance from the edge of each poster to the nearest edge of the wall is the same as the distance between the posters, as shown in the diagram below. Determine that distance.

𝟏𝟏𝟏𝟏 − 𝟏𝟏 − 𝟏𝟏 = 𝟔𝟔

𝟔𝟔 ÷ 𝟑𝟑 = 𝟐𝟐

The distance is 𝟐𝟐 feet.


Convey to students that the goal of this lesson is to learn how to build expressions and then write equations from those expressions.

First, using the fact that the distance between the wall and poster is two feet, write an expression (in terms of the distances between and the width of the posters) for the total length of the wall.

2 + 4 + 2 + 4 + 2 or 3(2) + 4 + 4 or 3(2) + 8

The numerical expressions you just wrote are based upon already knowing the answer. Suppose we wanted to solve the problem using algebra and did not know the answer is two feet. Let the distance between a picture and the nearest edge of the wall be 𝑥𝑥 feet. Write an expression for the total length of such a wall in terms of 𝑥𝑥. 𝑥𝑥 + 4 + 𝑥𝑥 + 4 + 𝑥𝑥 or 3𝑥𝑥 + 4 + 4 or 3𝑥𝑥 + 8

Setting this expression equal to the total length of the wall described in the problem, 14 feet, gives an equation. Write that equation.

𝑥𝑥 + 4 + 𝑥𝑥 + 4 + 𝑥𝑥 = 14 3𝑥𝑥 + 4 + 4 = 14

3𝑥𝑥 + 8 = 14

Using your answer from the Opening Exercise, check to see if your answer makes the equation true or false. Is the calculated distance consistent with the diagram that was drawn?

I got 2 as my answer in the Opening Exercise, so my equation is true.

My equation is false because I did not get 2 as my answer in the Opening Exercise.

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We say that 2 is a solution to the equation 3𝑥𝑥 + 8 = 14 because when it is substituted into the equation for 𝑥𝑥, it makes the equation a true number sentence: 3(2) + 8 = 14.

Your parents are redecorating the dining room and want to place two rectangular wall sconce lights that are 𝟐𝟐𝟐𝟐 inches

wide along a 𝟏𝟏𝟏𝟏𝟐𝟐𝟑𝟑-foot wall so that the distance between the lights and the distances from each light to the nearest edge of the wall are all the same. Design the wall and determine the distance.

𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢. = 𝟐𝟐𝟐𝟐𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟. = 𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟.

�𝟏𝟏𝟏𝟏𝟐𝟐𝟑𝟑

𝐟𝐟𝐟𝐟. − 𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟. − 𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟. � ÷ 𝟑𝟑

�𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟. − 𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟. − 𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟. � ÷ 𝟑𝟑

�𝟔𝟔𝟔𝟔𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟. � ÷ 𝟑𝟑

𝟔𝟔𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟. ÷ 𝟑𝟑

𝟏𝟏𝟑𝟑𝟐𝟐

𝐟𝐟𝐟𝐟. ÷ 𝟑𝟑

𝟏𝟏𝟑𝟑𝟐𝟐

𝐟𝐟𝐟𝐟. × 𝟏𝟏𝟑𝟑

𝟏𝟏𝟑𝟑𝟔𝟔

𝐟𝐟𝐟𝐟.

𝟐𝟐𝟏𝟏𝟔𝟔

𝐟𝐟𝐟𝐟.

OR

𝟏𝟏𝟏𝟏𝟐𝟐𝟑𝟑

𝐟𝐟𝐟𝐟.

𝟏𝟏𝟏𝟏 𝒇𝒇𝒇𝒇.× 𝟏𝟏𝟐𝟐𝐢𝐢𝐢𝐢𝐟𝐟𝐟𝐟 +

𝟐𝟐𝟑𝟑

𝐟𝐟𝐟𝐟. × 𝟏𝟏𝟐𝟐𝐢𝐢𝐢𝐢𝐟𝐟𝐟𝐟

𝟏𝟏𝟐𝟐𝟏𝟏 𝐢𝐢𝐢𝐢. + 𝟖𝟖 𝐢𝐢𝐢𝐢. 𝟏𝟏𝟐𝟐𝟖𝟖 𝐢𝐢𝐢𝐢.

(𝟏𝟏𝟐𝟐𝟖𝟖 𝐢𝐢𝐢𝐢. − 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢. − 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢. ) ÷ 𝟑𝟑 𝟕𝟕𝟖𝟖 𝐢𝐢𝐢𝐢. ÷ 𝟑𝟑 𝟐𝟐𝟔𝟔 𝐢𝐢𝐢𝐢.

Let the distance between a light and the nearest edge of a wall be 𝒙𝒙 𝐟𝐟𝐟𝐟. Write an expression in terms of 𝒙𝒙 for the total length of the wall. Then, use the expression and the length of the wall given in the problem to write an equation that can be used to find that distance.

𝟑𝟑𝒙𝒙 + 𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

+ 𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

𝟑𝟑𝒙𝒙 + 𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

+ 𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

= 𝟏𝟏𝟏𝟏𝟐𝟐𝟑𝟑

Scaffolding: Review that 12 inches = 1 foot.

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Now write an equation where 𝒚𝒚 stands for the number of inches: Let the distance between a light and the nearest edge of a wall be 𝒚𝒚 inches. Write an expression in terms of 𝒚𝒚 for the total length of the wall. Then, use the expression and the length of the wall to write an equation that can be used to find that distance (in inches).

𝟐𝟐 𝟏𝟏𝟏𝟏𝟐𝟐 feet = 𝟐𝟐𝟐𝟐 inches; therefore, the expression is 𝟑𝟑𝒚𝒚+ 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐.

𝟏𝟏𝟏𝟏𝟐𝟐𝟑𝟑 feet = 𝟏𝟏𝟐𝟐𝟖𝟖 inches; therefore, the equation is 𝟑𝟑𝒚𝒚+ 𝟐𝟐𝟐𝟐+ 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖.

What value(s) of 𝒚𝒚 makes the second equation true: 𝟐𝟐𝟏𝟏, 𝟐𝟐𝟐𝟐, or 𝟐𝟐𝟔𝟔?

𝒚𝒚 = 𝟐𝟐𝟏𝟏 𝟑𝟑𝒚𝒚 + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖

𝟑𝟑(𝟐𝟐𝟏𝟏) + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖 𝟕𝟕𝟐𝟐 + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖

𝟏𝟏𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖

False

𝒚𝒚 = 𝟐𝟐𝟐𝟐 𝟑𝟑𝒚𝒚 + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖

𝟑𝟑(𝟐𝟐𝟐𝟐) + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖 𝟕𝟕𝟐𝟐 + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖

𝟏𝟏𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖

False

𝒚𝒚 = 𝟐𝟐𝟔𝟔 𝟑𝟑𝒚𝒚 + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖

𝟑𝟑(𝟐𝟐𝟔𝟔) + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖 𝟕𝟕𝟖𝟖 + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟖𝟖

𝟏𝟏𝟐𝟐𝟖𝟖 = 𝟏𝟏𝟐𝟐𝟖𝟖

True

Since substituting 𝟐𝟐𝟔𝟔 for 𝒚𝒚 results in a true equation, the distance between the light and the nearest edge of the wall should be 𝟐𝟐𝟔𝟔 𝐢𝐢𝐢𝐢.


How did the change in the dimensions on the second problem change how you approached the problem?

The fractional width of 10 23 feet makes the arithmetic more difficult. The widths of the posters are

expressed in a different unit than the width of the room. This also makes the problem more difficult.

Since this problem is more difficult than the first, what additional steps are required to solve the problem?

The widths must be in the same units. Therefore, you must either convert 10 23 feet to inches or convert

25 inches to feet.

Describe the process of converting the units. To convert 25 inches to feet, divide the 25 inches by 12, and write the quotient as a mixed number. To

convert 10 23 feet to inches, multiply the whole number 10 by 12, and then multiply the fractional part,

23

, by 12; add the parts together to get the total number of inches.

After looking at both of the arithmetic solutions, which one seems the most efficient and why? Converting the width of the room from feet to inches made the overall problem shorter and easier

because after converting the width, all of the dimensions ended up being whole numbers and not fractions.

Does it matter which equation you use when determining which given values make the equation true? Explain how you know. Yes, since the values were given in inches, the equation 3𝑦𝑦 + 25 + 25 = 128 can be used because each

term of the equation is in the same unit of measure.

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7•3 Lesson 7


𝟏𝟏𝟐𝟐 𝟏𝟏

𝟏𝟏 𝟏𝟏

If one uses the other equation, what must be done to obtain the solution?

If the other equation were used, then the given values of 24, 25, and 26 inches need to be converted to

2, 2 112, and 2 1

6 feet, respectively.

Example (10 minutes)

The example is a consecutive integer word problem. A tape diagram is used to model an arithmetic solution in part (a). Replacing the first bar (the youngest sister’s age) in the tape diagram with 𝑥𝑥 years provides an opportunity for students to visualize the meaning of the equation created in part (b).

Example

The ages of three sisters are consecutive integers. The sum of their ages is 𝟏𝟏𝟐𝟐. Calculate their ages.

a. Use a tape diagram to find their ages.

Youngest sister

2nd sister

Oldest sister

𝟏𝟏𝟐𝟐 − 𝟑𝟑 = 𝟏𝟏𝟐𝟐

𝟏𝟏𝟐𝟐÷ 𝟑𝟑 = 𝟏𝟏𝟏𝟏

Youngest sister: 𝟏𝟏𝟏𝟏 years old

2nd sister: 𝟏𝟏𝟐𝟐 years old

Oldest sister: 𝟏𝟏𝟔𝟔 years old

b. If the youngest sister is 𝒙𝒙 years old, describe the ages of the other two sisters in terms of 𝒙𝒙, write an expression for the sum of their ages in terms of 𝒙𝒙, and use that expression to write an equation that can be used to find their ages.

Youngest sister: 𝒙𝒙 years old

2nd sister: (𝒙𝒙+ 𝟏𝟏) years old

Oldest sister: (𝒙𝒙+ 𝟐𝟐) years old

Sum of their ages: 𝒙𝒙 + (𝒙𝒙 + 𝟏𝟏) + (𝒙𝒙 + 𝟐𝟐)

Equation: 𝒙𝒙 + (𝒙𝒙 + 𝟏𝟏) + (𝒙𝒙 + 𝟐𝟐) = 𝟏𝟏𝟐𝟐

𝒙𝒙

𝒙𝒙

𝒙𝒙

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c. Determine if your answer from part (a) is a solution to the equation you wrote in part (b).

𝒙𝒙 + (𝒙𝒙 + 𝟏𝟏) + (𝒙𝒙 + 𝟐𝟐) = 𝟏𝟏𝟐𝟐 𝟏𝟏𝟏𝟏+ (𝟏𝟏𝟏𝟏+ 𝟏𝟏) + (𝟏𝟏𝟏𝟏+ 𝟐𝟐) = 𝟏𝟏𝟐𝟐

𝟏𝟏𝟐𝟐 = 𝟏𝟏𝟐𝟐

True

Let 𝑥𝑥 be an integer; write an algebraic expression that represents one more than that integer. 𝑥𝑥 + 1

Write an algebraic expression that represents two more than that integer. 𝑥𝑥 + 2

Discuss how the unknown unit in a tape diagram represents the unknown integer, represented by 𝑥𝑥. Consecutive integers begin with the unknown unit; then, every consecutive integer thereafter increases by 1 unit.

Exercise (8 minutes)

Instruct students to complete the following exercise individually and discuss the solution as a class.

Exercise

Sophia pays a $𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 membership fee for an online music store.

a. If she also buys two songs from a new album at a price of $𝟏𝟏.𝟏𝟏𝟏𝟏 each, what is the total cost?

$𝟐𝟐𝟏𝟏.𝟏𝟏𝟕𝟕

b. If Sophia purchases 𝒏𝒏 songs for $𝟏𝟏.𝟏𝟏𝟏𝟏 each, write an expression for the total cost.

𝟏𝟏.𝟏𝟏𝟏𝟏𝒏𝒏 + 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏

c. Sophia’s friend has saved $𝟏𝟏𝟏𝟏𝟖𝟖 but is not sure how many songs she can afford if she buys the membership and some songs. Use the expression in part (b) to write an equation that can be used to determine how many songs Sophia’s friend can buy.

𝟏𝟏.𝟏𝟏𝟏𝟏𝒏𝒏 + 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖

d. Using the equation written in part (c), can Sophia’s friend buy 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏𝟏𝟏, or 𝟏𝟏𝟏𝟏 songs?

𝒏𝒏 = 𝟏𝟏𝟏𝟏 𝟏𝟏.𝟏𝟏𝟏𝟏𝒏𝒏+ 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖

𝟏𝟏.𝟏𝟏𝟏𝟏(𝟏𝟏𝟏𝟏) + 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖 𝟏𝟏𝟖𝟖.𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖

𝟏𝟏𝟏𝟏𝟖𝟖 = 𝟏𝟏𝟏𝟏𝟖𝟖

True

𝒏𝒏 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏.𝟏𝟏𝟏𝟏𝒏𝒏+ 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖

𝟏𝟏.𝟏𝟏𝟏𝟏(𝟏𝟏𝟏𝟏𝟏𝟏) + 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖 𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖

𝟏𝟏𝟏𝟏𝟖𝟖.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖

False

𝒏𝒏 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏.𝟏𝟏𝟏𝟏𝒏𝒏+ 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖

𝟏𝟏.𝟏𝟏𝟏𝟏(𝟏𝟏𝟏𝟏𝟏𝟏) + 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏.𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟖𝟖

𝟏𝟏𝟏𝟏𝟏𝟏.𝟏𝟏𝟖𝟖 = 𝟏𝟏𝟏𝟏𝟖𝟖

False

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Closing (3 minutes)

Describe the process you used today to create an equation: What did you build first? What did you set it equal to?

Describe how to determine if a number is a solution to an equation.

Relevant Vocabulary

VARIABLE (DESCRIPTION): A variable is a symbol (such as a letter) that represents a number (i.e., it is a placeholder for a number).

EQUATION: An equation is a statement of equality between two expressions.

NUMBER SENTENCE: A number sentence is a statement of equality between two numerical expressions.

SOLUTION: A solution to an equation with one variable is a number that, when substituted for the variable in both expressions, makes the equation a true number sentence.


Lesson Summary

In many word problems, an equation is often formed by setting an expression equal to a number. To build the expression, it is helpful to consider a few numerical calculations with just numbers first. For example, if a pound of apples costs $𝟐𝟐, then three pounds cost $𝟔𝟔 (𝟐𝟐 × 𝟑𝟑), four pounds cost $𝟖𝟖 (𝟐𝟐 × 𝟏𝟏), and 𝒏𝒏 pounds cost 𝟐𝟐𝒏𝒏 dollars. If we had $𝟏𝟏𝟐𝟐 to spend on apples and wanted to know how many pounds we could buy, we can use the expression

𝟐𝟐𝒏𝒏 to write an equation, 𝟐𝟐𝒏𝒏 = 𝟏𝟏𝟐𝟐, which can then be used to find the answer: 𝟕𝟕𝟏𝟏𝟐𝟐 pounds.

To determine if a number is a solution to an equation, substitute the number into the equation for the variable (letter) and check to see if the resulting number sentence is true. If it is true, then the number is a solution to the

equation. For example, 𝟕𝟕𝟏𝟏𝟐𝟐 is a solution to 𝟐𝟐𝒏𝒏 = 𝟏𝟏𝟐𝟐 because 𝟐𝟐�𝟕𝟕𝟏𝟏𝟐𝟐� = 𝟏𝟏𝟐𝟐.

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Name Date


Exit Ticket 1. Check whether the given value of 𝑥𝑥 is a solution to the equation. Justify your answer.

a. 13

(𝑥𝑥 + 4) = 20 𝑥𝑥 = 48

b. 3𝑥𝑥 − 1 = 5𝑥𝑥 + 10 𝑥𝑥 = −5 12

2. The total cost of four pens and seven mechanical pencils is $13.25. The cost of each pencil is 75 cents.

a. Using an arithmetic approach, find the cost of a pen.

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b. Let the cost of a pen be 𝑝𝑝 dollars. Write an expression for the total cost of four pens and seven mechanical pencils in terms of 𝑝𝑝.

c. Write an equation that could be used to find the cost of a pen.

d. Determine a value for 𝑝𝑝 for which the equation you wrote in part (b) is true.

e. Determine a value for 𝑝𝑝 for which the equation you wrote in part (b) is false.

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1. Check whether the given value of 𝒙𝒙 is a solution to the equation. Justify your answer.

a. 𝟏𝟏𝟑𝟑

(𝒙𝒙 + 𝟏𝟏) = 𝟐𝟐𝟏𝟏 𝒙𝒙 = 𝟏𝟏𝟖𝟖

𝟏𝟏𝟑𝟑 (𝟏𝟏𝟖𝟖 + 𝟏𝟏) = 𝟐𝟐𝟏𝟏

𝟏𝟏𝟑𝟑 (𝟐𝟐𝟐𝟐) = 𝟐𝟐𝟏𝟏

𝟏𝟏𝟕𝟕𝟏𝟏𝟑𝟑 = 𝟐𝟐𝟏𝟏

False, 𝟏𝟏𝟖𝟖 is NOT a solution to 𝟏𝟏𝟑𝟑

(𝒙𝒙 + 𝟏𝟏) = 𝟐𝟐𝟏𝟏.

b. 𝟑𝟑𝒙𝒙 − 𝟏𝟏 = 𝟐𝟐𝒙𝒙 + 𝟏𝟏𝟏𝟏 𝒙𝒙 = −𝟐𝟐𝟏𝟏𝟐𝟐

𝟑𝟑�−𝟐𝟐𝟏𝟏𝟐𝟐� − 𝟏𝟏 = 𝟐𝟐�−𝟐𝟐𝟏𝟏𝟐𝟐� + 𝟏𝟏𝟏𝟏

−𝟑𝟑𝟑𝟑𝟐𝟐 − 𝟏𝟏 = −𝟐𝟐𝟐𝟐

𝟐𝟐 + 𝟏𝟏𝟏𝟏

−𝟑𝟑𝟐𝟐𝟐𝟐 = −𝟑𝟑𝟐𝟐

𝟐𝟐

True, −𝟐𝟐𝟏𝟏𝟐𝟐 is a solution to 𝟑𝟑𝒙𝒙 − 𝟏𝟏 = 𝟐𝟐𝒙𝒙 + 𝟏𝟏𝟏𝟏.

2. The total cost of four pens and seven mechanical pencils is $𝟏𝟏𝟑𝟑.𝟐𝟐𝟐𝟐. The cost of each pencil is 𝟕𝟕𝟐𝟐 cents.

a. Using an arithmetic approach, find the cost of a pen.

�𝟏𝟏𝟑𝟑.𝟐𝟐𝟐𝟐 − 𝟕𝟕(𝟏𝟏.𝟕𝟕𝟐𝟐)�÷ 𝟏𝟏

(𝟏𝟏𝟑𝟑.𝟐𝟐𝟐𝟐 − 𝟐𝟐.𝟐𝟐𝟐𝟐) ÷ 𝟏𝟏

𝟖𝟖 ÷ 𝟏𝟏

𝟐𝟐

The cost of a pen is $𝟐𝟐.

b. Let the cost of a pen be 𝒑𝒑 dollars. Write an expression for the total cost of four pens and seven mechanical pencils in terms of 𝒑𝒑.

𝟏𝟏𝒑𝒑 + 𝟕𝟕(𝟏𝟏.𝟕𝟕𝟐𝟐) or 𝟏𝟏𝒑𝒑 + 𝟐𝟐.𝟐𝟐𝟐𝟐

c. Write an equation that could be used to find the cost of a pen.

𝟏𝟏𝒑𝒑 + 𝟕𝟕(𝟏𝟏.𝟕𝟕𝟐𝟐) = 𝟏𝟏𝟑𝟑.𝟐𝟐𝟐𝟐 or 𝟏𝟏𝒑𝒑 + 𝟐𝟐.𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟑𝟑.𝟐𝟐𝟐𝟐

d. Determine a value for 𝒑𝒑 for which the equation you wrote in part (b) is true.

𝟏𝟏𝒑𝒑 + 𝟐𝟐.𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟑𝟑.𝟐𝟐𝟐𝟐𝟏𝟏(𝟐𝟐) + 𝟐𝟐.𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟑𝟑.𝟐𝟐𝟐𝟐

𝟖𝟖 + 𝟐𝟐.𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟑𝟑.𝟐𝟐𝟐𝟐𝟏𝟏𝟑𝟑.𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟑𝟑.𝟐𝟐𝟐𝟐

True, when 𝒑𝒑 = 𝟐𝟐, the equation is true.

e. Determine a value for 𝒑𝒑 for which the equation you wrote in part (b) is false.

Any value other than 𝟐𝟐 will make the equation false.

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1. Check whether the given value is a solution to the equation.

a. 𝟏𝟏𝒏𝒏 − 𝟑𝟑 = −𝟐𝟐𝒏𝒏 + 𝟏𝟏 𝒏𝒏 = 𝟐𝟐

𝟏𝟏(𝟐𝟐) − 𝟑𝟑 = −𝟐𝟐(𝟐𝟐) + 𝟏𝟏 𝟖𝟖 − 𝟑𝟑 = −𝟏𝟏 + 𝟏𝟏

𝟐𝟐 = 𝟐𝟐

True

b. 𝟏𝟏𝒎𝒎− 𝟏𝟏𝟏𝟏 = 𝟑𝟑𝒎𝒎+ 𝟏𝟏 𝒎𝒎 = 𝟏𝟏𝟏𝟏𝟑𝟑

𝟏𝟏�𝟏𝟏𝟏𝟏𝟑𝟑� − 𝟏𝟏𝟏𝟏 = 𝟑𝟑�

𝟏𝟏𝟏𝟏𝟑𝟑� + 𝟏𝟏

𝟏𝟏𝟏𝟏𝟑𝟑− 𝟏𝟏𝟏𝟏 =

𝟑𝟑𝟏𝟏𝟑𝟑

+ 𝟏𝟏

𝟑𝟑𝟏𝟏 − 𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏 + 𝟏𝟏 𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏

True

c. 𝟑𝟑(𝒚𝒚+ 𝟖𝟖) = 𝟐𝟐𝒚𝒚 − 𝟔𝟔 𝒚𝒚 = 𝟑𝟑𝟏𝟏

𝟑𝟑(𝟑𝟑𝟏𝟏+ 𝟖𝟖) = 𝟐𝟐(𝟑𝟑𝟏𝟏) − 𝟔𝟔 𝟑𝟑(𝟑𝟑𝟖𝟖) = 𝟔𝟔𝟏𝟏 − 𝟔𝟔 𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟐𝟐𝟏𝟏

False

2. Tell whether each number is a solution to the problem modeled by the following equation.

Mystery Number: Five more than −𝟖𝟖 times a number is 𝟐𝟐𝟏𝟏. What is the number?

Let the mystery number be represented by 𝒏𝒏.

The equation is 𝟐𝟐+ (−𝟖𝟖)𝒏𝒏 = 𝟐𝟐𝟏𝟏.

a. Is 𝟑𝟑 a solution to the equation? Why or why not?

No, because 𝟐𝟐 − 𝟐𝟐𝟏𝟏 ≠ 𝟐𝟐𝟏𝟏.

b. Is −𝟏𝟏 a solution to the equation? Why or why not?

No, because 𝟐𝟐 + 𝟑𝟑𝟐𝟐 ≠ 𝟐𝟐𝟏𝟏.

c. Is −𝟑𝟑 a solution to the equation? Why or why not?

Yes, because 𝟐𝟐+ 𝟐𝟐𝟏𝟏 = 𝟐𝟐𝟏𝟏.

d. What is the mystery number?

−𝟑𝟑 because 𝟐𝟐 more than −𝟖𝟖 times −𝟑𝟑 is 𝟐𝟐𝟏𝟏.

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3. The sum of three consecutive integers is 𝟑𝟑𝟔𝟔.

a. Find the smallest integer using a tape diagram.

1st integer

2nd integer

3rd integer

𝟑𝟑𝟔𝟔 − 𝟑𝟑 = 𝟑𝟑𝟑𝟑

𝟑𝟑𝟑𝟑÷ 𝟑𝟑 = 𝟏𝟏𝟏𝟏

The smallest integer is 𝟏𝟏𝟏𝟏.

b. Let 𝒏𝒏 represent the smallest integer. Write an equation that can be used to find the smallest integer.

Smallest integer: 𝒏𝒏

2nd integer: (𝒏𝒏+ 𝟏𝟏)

3rd integer: (𝒏𝒏+ 𝟐𝟐)

Sum of the three consecutive integers: 𝒏𝒏 + (𝒏𝒏+ 𝟏𝟏) + (𝒏𝒏+ 𝟐𝟐)

Equation: 𝒏𝒏 + (𝒏𝒏 + 𝟏𝟏) + (𝒏𝒏+ 𝟐𝟐) = 𝟑𝟑𝟔𝟔.

c. Determine if each value of 𝒏𝒏 below is a solution to the equation in part (b).

𝒏𝒏 = 𝟏𝟏𝟐𝟐.𝟐𝟐 No, it is not an integer and does not make a true equation.

𝒏𝒏 = 𝟏𝟏𝟐𝟐 No, it does not make a true equation.

𝒏𝒏 = 𝟏𝟏𝟏𝟏 Yes, it makes a true equation.

4. Andrew is trying to create a number puzzle for his younger sister to solve. He challenges his sister to find the mystery number. “When 𝟏𝟏 is subtracted from half of a number, the result is 𝟐𝟐.” The equation to represent the

mystery number is 𝟏𝟏𝟐𝟐𝒎𝒎 − 𝟏𝟏 = 𝟐𝟐. Andrew’s sister tries to guess the mystery number.

a. Her first guess is 𝟑𝟑𝟏𝟏. Is she correct? Why or why not?

No, it does not make a true equation. 𝟏𝟏𝟐𝟐

(𝟑𝟑𝟏𝟏) − 𝟏𝟏 = 𝟐𝟐

𝟏𝟏𝟐𝟐 − 𝟏𝟏 = 𝟐𝟐 𝟏𝟏𝟏𝟏 = 𝟐𝟐

False

b. Her second guess is 𝟐𝟐. Is she correct? Why or why not?

No, it does not make a true equation.

𝟏𝟏𝟐𝟐

(𝟐𝟐) − 𝟏𝟏 = 𝟐𝟐

𝟏𝟏 − 𝟏𝟏 = 𝟐𝟐 −𝟑𝟑 = 𝟐𝟐

False

𝟑𝟑𝟔𝟔 𝟏𝟏

𝟏𝟏 𝟏𝟏

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c. Her final guess is 𝟏𝟏𝟏𝟏𝟐𝟐. Is she correct? Why or why not?

No, it does not make a true equation.

𝟏𝟏𝟐𝟐�𝟏𝟏

𝟏𝟏𝟐𝟐� − 𝟏𝟏 = 𝟐𝟐

𝟐𝟐𝟏𝟏𝟏𝟏− 𝟏𝟏 = 𝟐𝟐

−𝟏𝟏𝟑𝟑𝟏𝟏

= 𝟐𝟐

False

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7•3 Lesson 8

Lesson 8: Using If-Then Moves in Solving Equations


Student Outcomes

Students understand and use the addition, subtraction, multiplication, division, and substitution properties of equality to solve word problems leading to equations of the form 𝑝𝑝𝑝𝑝 + 𝑞𝑞 = 𝑟𝑟 and 𝑝𝑝(𝑝𝑝 + 𝑞𝑞) = 𝑟𝑟 where 𝑝𝑝, 𝑞𝑞, and 𝑟𝑟 are specific rational numbers.

Students understand that any equation with rational coefficients can be written as an equation with expressions that involve only integer coefficients by multiplying both sides by the least common multiple of all the rational number terms.

Lesson Notes The intent of this lesson is for students to make the transition from an arithmetic approach of solving a word problem to an algebraic approach of solving the same problem. Recall from Module 2 that the process for solving linear equations is to isolate the variable by making 0s and 1s. In this module, the emphasis is for students to rewrite an equation using if-then moves into a form where the solution is easily recognizable. The main issue is that, in later grades, equations are rarely solved by isolating the variable (e.g., How do you isolate the variable for 3𝑝𝑝2 − 8 = −2𝑝𝑝?). Instead, students learn how to rewrite equations into different forms where the solutions are easy to recognize.

Examples of Grade 7 forms: The equation 23𝑝𝑝 + 27 = 31 is put into the form 𝑝𝑝 = 6, where it is easy to

recognize that the solution is 6.

Example of an Algebra I form: The equation 3𝑝𝑝2 − 8 = −2𝑝𝑝 is put into factored form (3𝑝𝑝 − 4)(𝑝𝑝 + 2) = 0,

where it is easy to recognize that the solutions are �43 ,−2�.

Example of an Algebra II and Precalculus form: The equation sin3 𝑝𝑝 + sin 𝑝𝑝 cos2 𝑝𝑝 = cos 𝑝𝑝 sin2 𝑝𝑝 + cos3 𝑝𝑝 is simplified to tan 𝑝𝑝 = 1, where it is easy to recognize that the solutions are �𝜋𝜋4 + 𝑘𝑘𝜋𝜋 � 𝑘𝑘 integer�.

Regardless of the type of equation students are studying, the if-then moves play an essential role in rewriting equations into different useful forms for solving, graphing, etc.

The FAQs on solving equations below are designed to help teachers understand the structure of the next set of lessons. Before reading the FAQ, it may be helpful to review the properties of operations and the properties of equality listed in Table 3 and Table 4 of the Common Core State Standards of Mathematics (CCSS-M).

What are the if-then moves? Recall the following if-then moves from Lesson 21 of Module 2:

1. Addition property of equality: If 𝑎𝑎 = 𝑏𝑏, then 𝑎𝑎 + 𝑐𝑐 = 𝑏𝑏 + 𝑐𝑐.

2. Subtraction property of equality: If 𝑎𝑎 = 𝑏𝑏, then 𝑎𝑎 − 𝑐𝑐 = 𝑏𝑏 − 𝑐𝑐. 3. Multiplication property of equality: If 𝑎𝑎 = 𝑏𝑏, then 𝑎𝑎 × 𝑐𝑐 = 𝑏𝑏 × 𝑐𝑐.

4. Division property of equality: If 𝑎𝑎 = 𝑏𝑏 and 𝑐𝑐 ≠ 0, then 𝑎𝑎 ÷ 𝑐𝑐 = 𝑏𝑏 ÷ 𝑐𝑐.

All eight properties of equality listed in Table 4 of the CCSS-M are if-then statements used in solving equations, but these four properties are separated out and collectively called the if-then moves.

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What points should I try to communicate to my students about solving equations? The goal is to make three important points about solving equations.

The technique for solving equations of the form 𝑝𝑝𝑝𝑝 + 𝑞𝑞 = 𝑟𝑟 and 𝑝𝑝(𝑝𝑝 + 𝑞𝑞) = 𝑟𝑟 is to rewrite them into the form 𝑝𝑝 = a number, using the properties of operations (Lessons 1–6) and the properties of equality (i.e., the if-then moves) to make 0s and 1s. This technique is sometimes called isolating the variable, but that name really only applies to linear equations. Consider mentioning that students will learn other techniques for other types of equations in later grades.

The properties of operations are used to modify one side of an equation at a time by changing the expression on a side into another equivalent expression.

The if-then moves are used to modify both sides of an equation simultaneously in a controlled way. The two expressions in the new equation are different than the two expressions in the old equation, but the solutions are the same.

How do if-then statements show up when solving equations? We will continue to use the normal convention of writing a sequence of equations underneath each other, linked together by if-then moves and/or properties of operations. For example, the sequence of equations and reasons for solving 3𝑝𝑝 = 3 is as follows:

13

(3𝑝𝑝) =13

(3) If-then move: multiply both sides by 13

.

�13 ⋅ 3� 𝑝𝑝 = 1

3 (3) Associative property

1 ⋅ 𝑝𝑝 = 1 Multiplicative inverse 𝑝𝑝 = 1 Multiplicative identity

This is a welcomed abbreviation for the if-then statements used in Lesson 21 of Module 2:

If 3𝑝𝑝 = 3, then 13

(3𝑝𝑝) =13

(3) by the if-then move of multiplying both sides by 13

.

If 13

(3𝑝𝑝) =13

(3), then �13 ⋅ 3� 𝑝𝑝 = 1

3 (3) by the associative property.

If �13 ⋅ 3� 𝑝𝑝 = 1

3 (3), then 1 ⋅ 𝑝𝑝 = 1 by the multiplicative inverse property.

If 1 ⋅ 𝑝𝑝 = 1, then 𝑝𝑝 = 1 by the multiplicative identity property.

The abbreviated form is visually much easier for students to understand provided that you explain to students that each pair of equations is part of an if-then statement.

In the unabbreviated if-then statements above, it looks like the properties of operations are also if-then statements. Are they? No. The properties of operations are not if-then statements themselves; most of them (associative, commutative, distributive, etc.) are statements about equivalent expressions. However, they are often used with combinations of the transitive and substitution properties of equality, which are if-then statements. For example, the transitive property states in this situation that if two expressions are equivalent, and if one of the expressions is substituted for the other in a true equation, then the resulting equation is also true (if 𝑎𝑎 = 𝑏𝑏 and 𝑏𝑏 = 𝑐𝑐, then 𝑎𝑎 = 𝑐𝑐).

Thus, the sentence above, If 13

(3𝑝𝑝) =13

(3), then (13 ⋅ 3)𝑝𝑝 = 1

3 (3) by the associative property, can be expanded as

follows:

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1. In solving the equation 13

(3𝑝𝑝) =13

(3), we assume 𝑝𝑝 is a number that makes this equation true.

2. �13 ⋅ 3� 𝑝𝑝 = 1

3 (3𝑝𝑝) is true by the associative property.

3. Therefore, we can replace the expression 13

(3𝑝𝑝) in the equation 13

(3𝑝𝑝) =13

(3) with the equivalent expression

�13 ⋅ 3� 𝑝𝑝 by the transitive property of equality.

(You might check that this fits the form of the transitive property described in the CCSS-M: If 𝑎𝑎 = 𝑏𝑏 and 𝑏𝑏 = 𝑐𝑐, then 𝑎𝑎 =𝑐𝑐.)

Teachers do not necessarily need to drill down to this level of detail when solving equations with students. Carefully monitor students for understanding and drill down to this level as needed.

Should I show every step in solving an equation? Yes and no: Please use your best judgment given the needs of your students. We generally do not write every step on the board when solving an equation. Otherwise, we would need to include discussions like the one above about the transitive property, which can throw off the lesson pace and detract from understanding. Here are general guidelines to follow when solving an equation with a class, which should work well with how these lessons are designed:

1. It is almost always better to initially include more steps than fewer. A good rule of thumb is to double the number of steps you would personally need to solve an equation. As adults, we do a lot more calculating in our heads than we realize. Doubling the number of steps slows down the pace of the lesson, which can be enormously beneficial to students.

2. As students catch on, look for ways to shorten the number of steps (for example, using any order/any grouping to collect all like terms at once rather than showing each associative/commutative property). Regardless, it is still important to verbally describe or ask for the properties being used in each step.

3. Write the reason (on the board) if it is one of the main concepts being learned in a lesson. For example, the next few lessons focus on if-then moves. Writing the if-then moves on the board calls them out to students and helps them focus on the main concept. As students become comfortable using the language of if-then moves, further reduce what you write on the board but verbally describe (or ask students to describe) the properties being used in each step.

We end with a quote from the High School, Algebra Progressions that encapsulates this entire FAQ:

“In the process of learning to solve equations, students learn certain ‘if-then’ moves, for example, ‘if 𝑝𝑝 = 𝑦𝑦, then 𝑝𝑝 + 2 = 𝑦𝑦 + 2.’ The danger in learning algebra is that students emerge with nothing but the moves, which may make it difficult to detect incorrect or made-up moves later on. Thus, the first requirement in the standards in this domain is that students understand that solving equations is a process of reasoning. This does not necessarily mean that they always write out the full text; part of the advantage of algebraic notation is its compactness. Once students know what the code stands for, they can start writing in code.”

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Classwork


Recall and summarize the if-then moves.

If a number is added or subtracted to both sides of a true equation, then the resulting equation is also true:

If 𝒂𝒂 = 𝒃𝒃, then 𝒂𝒂 + 𝒄𝒄 = 𝒃𝒃 + 𝒄𝒄.

If 𝒂𝒂 = 𝒃𝒃, then 𝒂𝒂 − 𝒄𝒄 = 𝒃𝒃 − 𝒄𝒄.

If a number is multiplied or divided to each side of a true equation, then the resulting equation is also true:

If 𝒂𝒂 = 𝒃𝒃, then 𝒂𝒂𝒄𝒄 = 𝒃𝒃𝒄𝒄.

If 𝒂𝒂 = 𝒃𝒃 and 𝒄𝒄 ≠ 𝟎𝟎, then 𝒂𝒂 ÷ 𝒄𝒄 = 𝒃𝒃 ÷ 𝒄𝒄.

Write 𝟑𝟑 + 𝟓𝟓 = 𝟖𝟖 in as many true equations as you can using the if-then moves. Identify which if-then move you used.

Answers will vary, but some examples are as follows:

If 𝟑𝟑 + 𝟓𝟓 = 𝟖𝟖, then 𝟑𝟑 + 𝟓𝟓+ 𝟒𝟒 = 𝟖𝟖+ 𝟒𝟒. Add 𝟒𝟒 to both sides.

If 𝟑𝟑 + 𝟓𝟓 = 𝟖𝟖, then 𝟑𝟑 + 𝟓𝟓 − 𝟒𝟒 = 𝟖𝟖 − 𝟒𝟒. Subtract 𝟒𝟒 from both sides.

If 𝟑𝟑 + 𝟓𝟓 = 𝟖𝟖, then 𝟒𝟒(𝟑𝟑 + 𝟓𝟓) = 𝟒𝟒(𝟖𝟖). Multiply both sides by 𝟒𝟒.

If 𝟑𝟑 + 𝟓𝟓 = 𝟖𝟖, then (𝟑𝟑 + 𝟓𝟓) ÷ 𝟒𝟒 = 𝟖𝟖 ÷ 𝟒𝟒. Divide both sides by 𝟒𝟒.


Example 1

Julia, Keller, and Israel are volunteer firefighters. On Saturday, the volunteer fire department held its annual coin drop fundraiser at a streetlight. After one hour, Keller had collected $𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎 more than Julia, and Israel had collected $𝟏𝟏𝟓𝟓 less than Keller. The three firefighters collected $𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 in total. How much did each person collect?

Find the solution using a tape diagram.

Julia

Keller

Israel

𝟑𝟑 units + 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎+ 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎 = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎+ 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎 = 𝟐𝟐𝟎𝟎

𝟑𝟑 units + 𝟐𝟐𝟎𝟎 = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 − 𝟐𝟐𝟎𝟎 = 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓

𝟑𝟑 units = 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓÷ 𝟑𝟑 = 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓

𝟏𝟏 unit = 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓

Julia collected $𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓. Keller collected $𝟔𝟔𝟏𝟏.𝟏𝟏𝟓𝟓. Israel collected $𝟒𝟒𝟔𝟔.𝟏𝟏𝟓𝟓.

What were the operations we used to get our answer?

First, we added 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎 and 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎 to get 𝟐𝟐𝟎𝟎. Next, we subtracted 𝟐𝟐𝟎𝟎 from 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓. Finally, we divided 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓 by 𝟑𝟑 to get 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓.

𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎

𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎 𝟏𝟏𝟓𝟓

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The amount of money Julia collected is 𝒋𝒋 dollars. Write an expression to represent the amount of money Keller collected in dollars.

𝒋𝒋 + 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎

Using the expressions for Julia and Keller, write an expression to represent the amount of money Israel collected in dollars.

𝒋𝒋 + 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎 − 𝟏𝟏𝟓𝟓

or

𝒋𝒋 + 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎

Using the expressions written above, write an equation in terms of 𝒋𝒋 that can be used to find the amount each person collected.

𝒋𝒋 + (𝒋𝒋 + 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎) + (𝒋𝒋+ 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎) = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓

Solve the equation written above to determine the amount of money each person collected, and describe any if-then moves used.

𝒋𝒋 + (𝒋𝒋 + 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎) + (𝒋𝒋+ 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎) = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓

𝟑𝟑𝒋𝒋 + 𝟐𝟐𝟎𝟎 = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 Any order, any grouping

(𝟑𝟑𝒋𝒋+ 𝟐𝟐𝟎𝟎) − 𝟐𝟐𝟎𝟎 = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 − 𝟐𝟐𝟎𝟎 If-then move: Subtract 𝟐𝟐𝟎𝟎 from both sides (to make a 𝟎𝟎).

𝟑𝟑𝒋𝒋 + 𝟎𝟎 = 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓 Any grouping, additive inverse

𝟑𝟑𝒋𝒋 = 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓 Additive identity

�𝟏𝟏𝟑𝟑� (𝟑𝟑𝒋𝒋) = (𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓) �𝟏𝟏𝟑𝟑� If-then move: Multiply both sides by 𝟏𝟏𝟑𝟑

(to make a 𝟏𝟏).

�𝟏𝟏𝟑𝟑 ⋅ 𝟑𝟑� 𝒋𝒋 = 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓 Associative property

𝟏𝟏 ⋅ 𝒋𝒋 = 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓 Multiplicative inverse

𝒋𝒋 = 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓 Multiplicative identity

If Julia collected $𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓, then Keller collected $𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓 + $𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎 = $𝟔𝟔𝟏𝟏.𝟏𝟏𝟓𝟓, and Israel collected

$𝟔𝟔𝟏𝟏.𝟏𝟏𝟓𝟓 − $𝟏𝟏𝟓𝟓 = $𝟒𝟒𝟔𝟔.𝟏𝟏𝟓𝟓.


Have students present the models they created based upon the given relationships, and then have the class compare different correct models and/or discuss why the incorrect models were incorrect. Some possible questions from the different models are as follows:

How does the tape diagram translate into the initial equation?

Each unknown unit represents how much Julia collected: 𝑗𝑗 dollars.

The initial step to solve the equation algebraically is to collect all like terms on the left-hand side of the equation using the any order, any grouping property.

The goal is to rewrite the equation into the form 𝑝𝑝 = a number by making zeros and ones.

Scaffolding: Teachers may need to review the process of solving an equation algebraically from Module 2, Lessons 17, 22, and 23.

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How can we make a zero or one?

Zeros are made with addition, and ones are made through multiplication and division.

We can make a 0 by subtracting 70 from both sides, or we can make a 1 by multiplying both sides by 13

.

(Both are correct if-then moves, but point out to students that making a 1 results in extra calculations.)

Let us subtract 70 from both sides. The if-then move of subtracting 70 from both sides changes both expressions of the equation (left and right sides) to new nonequivalent expressions, but the new expression has the same solution as the old one did.

In subtracting 70 from both sides, what do 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐 represent in the if-then move, If 𝑎𝑎 = 𝑏𝑏, then 𝑎𝑎 − 𝑐𝑐 = 𝑏𝑏 − 𝑐𝑐?

In this specific example, 𝑎𝑎 represents the left side of the equation, 3𝑗𝑗 + 70, 𝑏𝑏 represents the right side of the equation, 125.95, and 𝑐𝑐 is 70.

Continue to simplify the new equation using the properties of operations until reaching the equation 3𝑗𝑗 = 55.95. Can we make a zero or a one?

Yes, we can make a one by multiplying both sides by 13

. Since we are assuming that 𝑗𝑗 is a number that

makes the equation 3𝑗𝑗 = 55.95 true, we can apply the if-then move of multiplying both sides by 13

. The

resulting equation is also true.

How is the arithmetic approach (the tape diagram with arithmetic) similar to the algebraic approach (solving an equation)? The operations performed in solving the equation algebraically are the same operations done

arithmetically.

How can the equation 3𝑗𝑗 + 70 = 125.95 be written so that the equation contains only integers? What would the new equation be?

You can multiply each term by 100. The equivalent equation would be 300𝑗𝑗 + 7000 = 12595.

Show students that solving this problem also leads to 𝑗𝑗 = 18.65.

What if, instead, we used the amount Keller collected: 𝑘𝑘 dollars. Would that be okay? How would the money collected by the other people then be defined?

Yes, that would be okay. Since Keller has $42.50 more than Julia, then Julia would have $42.50 less than Keller. Julia’s money would be 𝑘𝑘 − 42.50. Since Israel’s money is $15.00 less than Keller, his money is 𝑘𝑘 − 15.

The expressions defining each person’s amount differ depending on who we choose to represent the other two people. Complete the chart to show how the statements vary when 𝑝𝑝 changes.

In terms of Julia Israel Keller Julia’s amount (𝑗𝑗) 𝑗𝑗 𝑗𝑗 + 27.50 𝑗𝑗 + 42 Israel’s amount (𝑖𝑖) 𝑖𝑖 − 27.50 𝑖𝑖 𝑖𝑖 + 15 Keller’s amount (𝑘𝑘) 𝑘𝑘 − 42.50 𝑘𝑘 − 15 𝑘𝑘

If time, set up and solve the equation in terms of 𝑘𝑘. Show students that the equation and solution are different than the equation based upon Julia’s amount, but that the solution, $61.15, matches how much Keller collected.

MP.2

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Example 2

You are designing a rectangular pet pen for your new baby puppy. You have 𝟑𝟑𝟎𝟎 feet of fence barrier. You decide that you

would like the length to be 𝟔𝟔𝟏𝟏𝟑𝟑 feet longer than the width.

Draw and label a diagram to represent the pet pen. Write expressions to represent the width and length of the pet pen.

Width of the pet pen: 𝒙𝒙 𝐟𝐟𝐟𝐟.

Then, �𝒙𝒙 + 𝟔𝟔𝟏𝟏𝟑𝟑� 𝐟𝐟𝐟𝐟. represents the length of the pet pen.

Find the dimensions of the pet pen.

Arithmetic Algebraic

�𝟑𝟑𝟎𝟎 − 𝟔𝟔𝟏𝟏𝟑𝟑− 𝟔𝟔

𝟏𝟏𝟑𝟑�÷ 𝟒𝟒

𝟏𝟏𝟐𝟐𝟏𝟏𝟑𝟑

÷ 𝟒𝟒

𝟒𝟒𝟏𝟏𝟑𝟑

The width is 𝟒𝟒𝟏𝟏𝟑𝟑 𝐟𝐟𝐟𝐟.

The length is 𝟒𝟒𝟏𝟏𝟑𝟑 𝐟𝐟𝐟𝐟. +𝟔𝟔𝟏𝟏𝟑𝟑 𝐟𝐟𝐟𝐟. = 𝟏𝟏𝟎𝟎𝟒𝟒𝟑𝟑 𝐟𝐟𝐟𝐟.

𝒙𝒙 + �𝒙𝒙 + 𝟔𝟔𝟏𝟏𝟑𝟑� + 𝒙𝒙 + �𝒙𝒙 + 𝟔𝟔

𝟏𝟏𝟑𝟑� = 𝟑𝟑𝟎𝟎

𝟒𝟒𝒙𝒙 + 𝟏𝟏𝟒𝟒𝟒𝟒𝟑𝟑

= 𝟑𝟑𝟎𝟎

𝟒𝟒𝒙𝒙 + 𝟏𝟏𝟒𝟒𝟒𝟒𝟑𝟑− 𝟏𝟏𝟒𝟒

𝟒𝟒𝟑𝟑

= 𝟑𝟑𝟎𝟎 − 𝟏𝟏𝟒𝟒𝟒𝟒𝟑𝟑

𝟒𝟒𝒙𝒙 = 𝟏𝟏𝟐𝟐𝟏𝟏𝟑𝟑

�𝟏𝟏𝟒𝟒� (𝟒𝟒𝒙𝒙) = �𝟏𝟏𝟐𝟐

𝟏𝟏𝟑𝟑� �𝟏𝟏𝟒𝟒�

𝒙𝒙 = 𝟒𝟒𝟏𝟏𝟑𝟑

If-then move: Subtract

𝟏𝟏𝟒𝟒𝟒𝟒𝟑𝟑 from both sides.

If-then move: Multiply

both sides by 𝟏𝟏𝟒𝟒

.

If the perimeter of the pet pen is 𝟑𝟑𝟎𝟎 𝐟𝐟𝐟𝐟. and the length of the pet pen is 𝟔𝟔𝟏𝟏𝟑𝟑 𝐟𝐟𝐟𝐟. longer than the width, then the width

would be 𝟒𝟒𝟏𝟏𝟑𝟑 𝐟𝐟𝐟𝐟., and the length would be 𝟒𝟒𝟏𝟏𝟑𝟑 𝐟𝐟𝐟𝐟. +𝟔𝟔𝟏𝟏𝟑𝟑 𝐟𝐟𝐟𝐟. = 𝟏𝟏𝟎𝟎𝟒𝟒𝟑𝟑 𝐟𝐟𝐟𝐟.

If an arithmetic approach was used to determine the dimensions, write an equation that can be used to find the dimensions. Encourage students to verbalize their strategy and the if-then moves used to rewrite the equation with the same solution.


Example 3

Nancy’s morning routine involves getting dressed, eating breakfast, making her bed, and driving to work. Nancy spends 𝟏𝟏𝟑𝟑

of the total time in the morning getting dressed, 𝟏𝟏𝟎𝟎 minutes eating breakfast, 𝟓𝟓 minutes making her bed, and the

remaining time driving to work. If Nancy spends 𝟑𝟑𝟓𝟓𝟏𝟏𝟒𝟒 minutes getting dressed, eating breakfast, and making her bed, how long is her drive to work?

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Write and solve this problem using an equation. Identify the if-then moves used when solving the equation.

Total time of routine: 𝒙𝒙 minutes

𝟏𝟏𝟑𝟑𝒙𝒙 + 𝟏𝟏𝟎𝟎 + 𝟓𝟓 = 𝟑𝟑𝟓𝟓

𝟏𝟏𝟒𝟒

𝟏𝟏𝟑𝟑𝒙𝒙 + 𝟏𝟏𝟓𝟓 = 𝟑𝟑𝟓𝟓

𝟏𝟏𝟒𝟒

𝟏𝟏𝟑𝟑𝒙𝒙 + 𝟏𝟏𝟓𝟓 − 𝟏𝟏𝟓𝟓 = 𝟑𝟑𝟓𝟓

𝟏𝟏𝟒𝟒− 𝟏𝟏𝟓𝟓

𝟏𝟏𝟑𝟑𝒙𝒙 + 𝟎𝟎 = 𝟒𝟒𝟎𝟎

𝟏𝟏𝟒𝟒

𝟑𝟑�𝟏𝟏𝟑𝟑𝒙𝒙� = 𝟑𝟑�𝟒𝟒𝟎𝟎

𝟏𝟏𝟒𝟒�

𝒙𝒙 = 𝟔𝟔𝟏𝟏𝟏𝟏𝟒𝟒

𝟔𝟔𝟏𝟏𝟏𝟏𝟒𝟒− 𝟑𝟑𝟓𝟓

𝟏𝟏𝟒𝟒

= 𝟒𝟒𝟔𝟔

It takes Nancy 𝟒𝟒𝟔𝟔 minutes to drive to work.

If-then move: Subtract 𝟏𝟏𝟓𝟓 from both sides.

If-then move: Multiply both sides by 𝟑𝟑.

Is your answer reasonable? Explain.

Yes, the answer is reasonable because some of the morning activities take 𝟑𝟑𝟓𝟓𝟏𝟏𝟒𝟒 minutes, so the total amount of time for

everything will be more than 𝟑𝟑𝟓𝟓𝟏𝟏𝟒𝟒 minutes. Also, when checking the total time for all of the morning routine, the total sum is equal to total time found. However, to find the time for driving to work, a specific activity in the morning, it is necessary to find the difference from the total time and all the other activities.

Encourage students to verbalize their strategy of solving the problem by identifying what the unknown represents and then using if-then moves to make 0 and 1.

What does the variable 𝑝𝑝 represent in the equation? 𝑝𝑝 represents the total amount of time of Nancy’s entire morning routine.

Explain how to use the answer for 𝑝𝑝 to determine the time that Nancy spends driving to work.

Since 𝑝𝑝 represents the total amont of time in the morning, and the problem gives the amount of time spent on all other activities besides driving, the total time spent driving is the difference of the two amounts. Therefore, you must subtract the total time and the time doing the other activities.

Discuss how the arithmetic approach can be modeled with a bar model.

MP.4

If-then move: Subtract 𝟏𝟏𝟓𝟓 from both sides.

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Getting dressed represents 13

of the total time as modeled.

We know part of the other morning activities takes a total of 15 minutes; therefore, part of a bar is drawn to model the 15 minutes.

We know that the bar that represents the time getting dressed and the other activities of 15 minutes equals a

total of 35 12 minutes. Therefore, the getting dressed bar is equal to 35 1

2 − 15 = 20 12.

The remaining bars that represent a third of the total time also equal 20 12. Therefore, the total time is

20 12 + 20 1

2 + 20 12 = 61 1

2.

The time spent driving would be equal to the total time less the time spent doing all other activities,

61 12 − 35 1

2 = 26.


Example 4

The total number of participants who went on the seventh-grade field trip to the Natural Science Museum consisted of all of the seventh-grade students and 𝟐𝟐 adult chaperones. Two-thirds of the total participants rode a large bus, and the rest rode a smaller bus. If 𝟓𝟓𝟒𝟒 students rode the large bus, how many students went on the field trip?

Arithmetic Approach:

Total on both buses: (𝟓𝟓𝟒𝟒÷ 𝟒𝟒) × 𝟑𝟑 = 𝟖𝟖𝟏𝟏

Total number of students: 𝟖𝟖𝟏𝟏 − 𝟐𝟐 = 𝟐𝟐𝟒𝟒; 𝟐𝟐𝟒𝟒 students went on the field trip.

Algebraic Approach: Challenge students to build the equation and solve it on their own first. Then, go through the steps with them, pointing out how we are “making zeros” and “making ones.” Point out that, in this problem, it is advantageous to make a 𝟏𝟏 first. (This example is an equation of the form 𝒑𝒑(𝒙𝒙+ 𝒒𝒒) = 𝒓𝒓.)

Number of students: 𝒔𝒔

Total number of participants: 𝒔𝒔 + 𝟐𝟐

𝟒𝟒𝟑𝟑

(𝒔𝒔 + 𝟐𝟐) = 𝟓𝟓𝟒𝟒

𝟑𝟑𝟒𝟒�

𝟒𝟒𝟑𝟑

(𝒔𝒔 + 𝟐𝟐)� =𝟑𝟑𝟒𝟒

(𝟓𝟓𝟒𝟒)

�𝟑𝟑𝟒𝟒⋅𝟒𝟒𝟑𝟑� (𝒔𝒔 + 𝟐𝟐) = 𝟖𝟖𝟏𝟏

𝟏𝟏(𝒔𝒔 + 𝟐𝟐) = 𝟖𝟖𝟏𝟏 𝒔𝒔 + 𝟐𝟐 = 𝟖𝟖𝟏𝟏

(𝒔𝒔 + 𝟐𝟐) − 𝟐𝟐 = 𝟖𝟖𝟏𝟏 − 𝟐𝟐 𝒔𝒔 + 𝟎𝟎 = 𝟐𝟐𝟒𝟒

𝒔𝒔 = 𝟐𝟐𝟒𝟒

𝟐𝟐𝟒𝟒 students went on the field trip.

If-then move: Multiply both sides by 𝟑𝟑𝟒𝟒

(to make a 𝟏𝟏).

If-then move: Subtract 𝟐𝟐 from both sides (to make a 𝟎𝟎).

𝟒𝟒𝟑𝟑 on large bus

𝟏𝟏𝟑𝟑

on smaller bus

𝟓𝟓𝟒𝟒 𝟐𝟐

MP.4

𝒔𝒔

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How can the model be used to write an equation?

By replacing the question mark with 𝑠𝑠, we see that the total number of participants is 𝑠𝑠 + 7. Since the

diagram shows that 23

of the total is 54, we can write 23

(𝑠𝑠 + 7) = 54.

How is the calculation (54 ÷ 2) × 3 in the arithmetic approach similar to making a 1 in the algebraic approach?

Dividing by 2 and multiplying by 3 is the same as multiplying by 32

.

Which approach did you prefer? Why?

Answers will vary, but try to bring out: The tape diagram in this problem was harder to construct than usual while the equation seemed to make more sense.

Closing (3 minutes)

Describe how if-then moves are applied to solving a word problem algebraically.

Compare the algebraic and arithmetic approaches. Name the similarities between them. Which approach do you prefer? Why?

How can equations be rewritten so the equation contains only integer coefficients and constants?


Lesson Summary

Algebraic Approach: To solve an equation algebraically means to use the properties of operations and if-then moves to simplify the equation into a form where the solution is easily recognizable. For the equations we are studying this year (called linear equations), that form is an equation that looks like 𝒙𝒙 = a number, where the number is the solution.

If-Then Moves: If 𝒙𝒙 is a solution to an equation, it will continue to be a solution to the new equation formed by adding or subtracting a number from both sides of the equation. It will also continue to be a solution when both sides of the equation are multiplied by or divided by a nonzero number. We use these if-then moves to make zeros and ones in ways that simplify the original equation.

Useful First Step: If one is faced with the task of finding a solution to an equation, a useful first step is to collect like terms on each side of the equation.

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7•3 Lesson 8


Name Date


Exit Ticket Mrs. Canale’s class is selling frozen pizzas to earn money for a field trip. For every pizza sold, the class makes $5.35. They have already earned $182.90 toward their $750 goal. How many more pizzas must they sell to earn $750? Solve this problem first by using an arithmetic approach, then by using an algebraic approach. Compare the calculations you made using each approach.

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Mrs. Canale’s class is selling frozen pizzas to earn money for a field trip. For every pizza sold, the class makes $𝟓𝟓.𝟑𝟑𝟓𝟓. They have already earned $𝟏𝟏𝟖𝟖𝟒𝟒.𝟗𝟗𝟎𝟎, but they need $𝟐𝟐𝟓𝟓𝟎𝟎. How many more pizzas must they sell to earn $𝟐𝟐𝟓𝟓𝟎𝟎? Solve this problem first by using an arithmetic approach, then by using an algebraic approach. Compare the calculations you made using each approach.

Arithmetic Approach:

Amount of money needed: 𝟐𝟐𝟓𝟓𝟎𝟎 − 𝟏𝟏𝟖𝟖𝟒𝟒.𝟗𝟗𝟎𝟎 = 𝟓𝟓𝟔𝟔𝟐𝟐.𝟏𝟏𝟎𝟎

Number of pizzas needed: 𝟓𝟓𝟔𝟔𝟐𝟐.𝟏𝟏𝟎𝟎 ÷ 𝟓𝟓.𝟑𝟑𝟓𝟓 = 𝟏𝟏𝟎𝟎𝟔𝟔

If the class wants to earn a total of $𝟐𝟐𝟓𝟓𝟎𝟎, then they must sell 𝟏𝟏𝟎𝟎𝟔𝟔 more pizzas.

Algebraic Approach:

Let 𝒙𝒙 represent the number of additional pizzas they need to sell.

𝟓𝟓.𝟑𝟑𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟖𝟖𝟒𝟒.𝟗𝟗𝟎𝟎 = 𝟐𝟐𝟓𝟓𝟎𝟎 𝟓𝟓.𝟑𝟑𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟖𝟖𝟒𝟒.𝟗𝟗𝟎𝟎 − 𝟏𝟏𝟖𝟖𝟒𝟒.𝟗𝟗𝟎𝟎 = 𝟐𝟐𝟓𝟓𝟎𝟎 − 𝟏𝟏𝟖𝟖𝟒𝟒.𝟗𝟗𝟎𝟎

𝟓𝟓.𝟑𝟑𝟓𝟓𝒙𝒙 + 𝟎𝟎 = 𝟓𝟓𝟔𝟔𝟐𝟐.𝟏𝟏𝟎𝟎

�𝟏𝟏

𝟓𝟓.𝟑𝟑𝟓𝟓� (𝟓𝟓.𝟑𝟑𝟓𝟓𝒙𝒙) = �

𝟏𝟏𝟓𝟓.𝟑𝟑𝟓𝟓

� (𝟓𝟓𝟔𝟔𝟐𝟐.𝟏𝟏𝟎𝟎)

𝒙𝒙 = 𝟏𝟏𝟎𝟎𝟔𝟔

OR

𝟓𝟓.𝟑𝟑𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟖𝟖𝟒𝟒.𝟗𝟗𝟎𝟎 = 𝟐𝟐𝟓𝟓𝟎𝟎 𝟏𝟏𝟎𝟎𝟎𝟎(𝟓𝟓.𝟑𝟑𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟖𝟖𝟒𝟒.𝟗𝟗𝟎𝟎) = 𝟏𝟏𝟎𝟎𝟎𝟎(𝟐𝟐𝟓𝟓𝟎𝟎)

𝟓𝟓𝟑𝟑𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟖𝟖𝟒𝟒𝟗𝟗𝟎𝟎 = 𝟐𝟐𝟓𝟓𝟎𝟎𝟎𝟎𝟎𝟎 𝟓𝟓𝟑𝟑𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟖𝟖𝟒𝟒𝟗𝟗𝟎𝟎 − 𝟏𝟏𝟖𝟖𝟒𝟒𝟗𝟗𝟎𝟎 = 𝟐𝟐𝟓𝟓𝟎𝟎𝟎𝟎𝟎𝟎 − 𝟏𝟏𝟖𝟖𝟒𝟒𝟗𝟗𝟎𝟎

�𝟏𝟏𝟓𝟓𝟑𝟑𝟓𝟓

� (𝟓𝟓𝟑𝟑𝟓𝟓𝒙𝒙) = �𝟏𝟏𝟓𝟓𝟑𝟑𝟓𝟓

� (𝟓𝟓𝟔𝟔𝟐𝟐𝟏𝟏𝟎𝟎)

𝒙𝒙 = 𝟏𝟏𝟎𝟎𝟔𝟔

If the class wants to earn $𝟐𝟐𝟓𝟓𝟎𝟎, then they must sell 𝟏𝟏𝟎𝟎𝟔𝟔 more pizzas.

Both approaches subtract 𝟏𝟏𝟖𝟖𝟒𝟒.𝟗𝟗𝟎𝟎 from 𝟐𝟐𝟓𝟓𝟎𝟎 to get 𝟓𝟓𝟔𝟔𝟐𝟐.𝟏𝟏𝟎𝟎. Dividing by 𝟓𝟓.𝟑𝟑𝟓𝟓 is the same as multiplying by 𝟏𝟏

𝟓𝟓.𝟑𝟑𝟓𝟓. Both

result in 𝟏𝟏𝟎𝟎𝟔𝟔 more pizzas that the class needs to sell.


Write and solve an equation for each problem.

1. The perimeter of a rectangle is 𝟑𝟑𝟎𝟎 inches. If its length is three times its width, find the dimensions.

The width of the rectangle: 𝒘𝒘 inches The length of the rectangle: 𝟑𝟑𝒘𝒘 inches

𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐏𝐟𝐟𝐏𝐏𝐏𝐏 = 𝟒𝟒(𝐥𝐥𝐏𝐏𝐥𝐥𝐥𝐥𝐟𝐟𝐥𝐥+ 𝐰𝐰𝐏𝐏𝐰𝐰𝐟𝐟𝐥𝐥)

𝟒𝟒(𝒘𝒘 + 𝟑𝟑𝒘𝒘) = 𝟑𝟑𝟎𝟎 𝟒𝟒(𝟒𝟒𝒘𝒘) = 𝟑𝟑𝟎𝟎

𝟖𝟖𝒘𝒘 = 𝟑𝟑𝟎𝟎

�𝟏𝟏𝟖𝟖� (𝟖𝟖𝒘𝒘) = �

𝟏𝟏𝟖𝟖� (𝟑𝟑𝟎𝟎)

𝒘𝒘 = 𝟑𝟑𝟑𝟑𝟒𝟒

OR

𝟒𝟒(𝒘𝒘 + 𝟑𝟑𝒘𝒘) = 𝟑𝟑𝟎𝟎 (𝒘𝒘 + 𝟑𝟑𝒘𝒘) = 𝟏𝟏𝟓𝟓

𝟒𝟒𝒘𝒘 = 𝟏𝟏𝟓𝟓

𝒘𝒘 = 𝟑𝟑𝟑𝟑𝟒𝟒

The width is 𝟑𝟑𝟑𝟑𝟒𝟒 inches.

The length is (𝟑𝟑) �𝟑𝟑𝟑𝟑𝟒𝟒 𝐏𝐏𝐥𝐥. � = (𝟑𝟑) �𝟏𝟏𝟓𝟓𝟒𝟒 𝐏𝐏𝐥𝐥. � = 𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒 𝐏𝐏𝐥𝐥.

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2. A cell phone company has a basic monthly plan of $𝟒𝟒𝟎𝟎 plus $𝟎𝟎.𝟒𝟒𝟓𝟓 for any minutes used over 𝟐𝟐𝟎𝟎𝟎𝟎. Before receiving his statement, John saw he was charged a total of $𝟒𝟒𝟖𝟖.𝟏𝟏𝟎𝟎. Write and solve an equation to determine how many minutes he must have used during the month. Write an equation without decimals.

The number of minutes over 𝟐𝟐𝟎𝟎𝟎𝟎: 𝒎𝒎 minutes

𝟒𝟒𝟎𝟎+ 𝟎𝟎.𝟒𝟒𝟓𝟓𝒎𝒎 = 𝟒𝟒𝟖𝟖.𝟏𝟏𝟎𝟎 𝟎𝟎.𝟒𝟒𝟓𝟓𝒎𝒎+ 𝟒𝟒𝟎𝟎 − 𝟒𝟒𝟎𝟎 = 𝟒𝟒𝟖𝟖.𝟏𝟏𝟎𝟎 − 𝟒𝟒𝟎𝟎

𝟎𝟎.𝟒𝟒𝟓𝟓𝒎𝒎 = 𝟖𝟖.𝟏𝟏𝟎𝟎

�𝟏𝟏

𝟎𝟎.𝟒𝟒𝟓𝟓� (𝟎𝟎.𝟒𝟒𝟓𝟓𝒎𝒎) = 𝟖𝟖.𝟏𝟏𝟎𝟎�

𝟏𝟏𝟎𝟎.𝟒𝟒𝟓𝟓

�

𝒎𝒎 = 𝟏𝟏𝟖𝟖

𝟒𝟒𝟎𝟎𝟎𝟎𝟎𝟎+ 𝟒𝟒𝟓𝟓𝒎𝒎 = 𝟒𝟒𝟖𝟖𝟏𝟏𝟎𝟎 𝟒𝟒𝟓𝟓𝒎𝒎+ 𝟒𝟒𝟎𝟎𝟎𝟎𝟎𝟎 − 𝟒𝟒𝟎𝟎𝟎𝟎𝟎𝟎 = 𝟒𝟒𝟖𝟖𝟏𝟏𝟎𝟎 − 𝟒𝟒𝟎𝟎𝟎𝟎𝟎𝟎

𝟒𝟒𝟓𝟓𝒎𝒎 = 𝟖𝟖𝟏𝟏𝟎𝟎

�𝟏𝟏𝟒𝟒𝟓𝟓� (𝟒𝟒𝟓𝟓𝒎𝒎) = 𝟖𝟖𝟏𝟏𝟎𝟎�

𝟏𝟏𝟒𝟒𝟓𝟓�

𝒎𝒎 = 𝟏𝟏𝟖𝟖

John used 𝟏𝟏𝟖𝟖 minutes over 𝟐𝟐𝟎𝟎𝟎𝟎 for the month. He used a total of 𝟐𝟐𝟏𝟏𝟖𝟖 minutes.

3. A volleyball coach plans her daily practices to include 𝟏𝟏𝟎𝟎 minutes of stretching, 𝟒𝟒𝟑𝟑

of the entire practice scrimmaging,

and the remaining practice time working on drills of specific skills. On Wednesday, the coach planned 𝟏𝟏𝟎𝟎𝟎𝟎 minutes of stretching and scrimmaging. How long, in hours, is the entire practice?

The duration of the entire practice: 𝒙𝒙 hours

𝟒𝟒𝟑𝟑𝒙𝒙 +

𝟏𝟏𝟎𝟎𝟔𝟔𝟎𝟎

=𝟏𝟏𝟎𝟎𝟎𝟎𝟔𝟔𝟎𝟎

𝟒𝟒𝟑𝟑𝒙𝒙 +

𝟏𝟏𝟔𝟔

=𝟓𝟓𝟑𝟑

𝟒𝟒𝟑𝟑𝒙𝒙 +

𝟏𝟏𝟔𝟔−𝟏𝟏𝟔𝟔

=𝟓𝟓𝟑𝟑−𝟏𝟏𝟔𝟔

𝟒𝟒𝟑𝟑𝒙𝒙 =

𝟗𝟗𝟔𝟔

�𝟑𝟑𝟒𝟒� �𝟒𝟒𝟑𝟑𝒙𝒙� =

𝟑𝟑𝟒𝟒�𝟗𝟗𝟔𝟔�

𝒙𝒙 =𝟒𝟒𝟐𝟐𝟏𝟏𝟒𝟒

= 𝟒𝟒𝟏𝟏𝟒𝟒

The entire practice is a length of 𝟒𝟒𝟏𝟏𝟒𝟒 hours, or 𝟒𝟒.𝟒𝟒𝟓𝟓 hours.

4. The sum of two consecutive even numbers is 𝟓𝟓𝟒𝟒. Find the numbers.

First consecutive even integer: 𝒙𝒙

Second consecutive even integer: 𝒙𝒙 + 𝟒𝟒

𝒙𝒙 + (𝒙𝒙 + 𝟒𝟒) = 𝟓𝟓𝟒𝟒 𝟒𝟒𝒙𝒙 + 𝟒𝟒 = 𝟓𝟓𝟒𝟒

𝟒𝟒𝒙𝒙 + 𝟒𝟒 − 𝟒𝟒 = 𝟓𝟓𝟒𝟒 − 𝟒𝟒 𝟒𝟒𝒙𝒙 + 𝟎𝟎 = 𝟓𝟓𝟒𝟒

�𝟏𝟏𝟒𝟒� (𝟒𝟒𝒙𝒙) = �

𝟏𝟏𝟒𝟒� (𝟓𝟓𝟒𝟒)

𝒙𝒙 = 𝟒𝟒𝟔𝟔

The consecutive even integers are 𝟒𝟒𝟔𝟔 and 𝟒𝟒𝟖𝟖.

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5. Justin has $𝟐𝟐.𝟓𝟓𝟎𝟎 more than Eva, and Emma has $𝟏𝟏𝟒𝟒 less than Justin. Together, they have a total of $𝟔𝟔𝟑𝟑.𝟎𝟎𝟎𝟎. How much money does each person have?

The amount of money Eva has: 𝒙𝒙 dollars

The amount of money Justin has: (𝒙𝒙 + 𝟐𝟐.𝟓𝟓𝟎𝟎) dollars

The amount of money Emma has: �(𝒙𝒙 + 𝟐𝟐.𝟓𝟓𝟎𝟎) − 𝟏𝟏𝟒𝟒� dollars, or (𝒙𝒙 − 𝟒𝟒.𝟓𝟓𝟎𝟎) dollars

𝒙𝒙 + (𝒙𝒙 + 𝟐𝟐.𝟓𝟓𝟎𝟎) + (𝒙𝒙 − 𝟒𝟒.𝟓𝟓𝟎𝟎) = 𝟔𝟔𝟑𝟑 𝟑𝟑𝒙𝒙 + 𝟑𝟑 = 𝟔𝟔𝟑𝟑

𝟑𝟑𝒙𝒙 + 𝟑𝟑 − 𝟑𝟑 = 𝟔𝟔𝟑𝟑 − 𝟑𝟑 𝟑𝟑𝒙𝒙 + 𝟎𝟎 = 𝟔𝟔𝟎𝟎

�𝟏𝟏𝟑𝟑�𝟑𝟑𝒙𝒙 = �

𝟏𝟏𝟑𝟑�𝟔𝟔𝟎𝟎

𝒙𝒙 = 𝟒𝟒𝟎𝟎

If the total amount of money all three people have is $𝟔𝟔𝟑𝟑, then Eva has $𝟒𝟒𝟎𝟎, Justin has $𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎, and Emma has $𝟏𝟏𝟓𝟓.𝟓𝟓𝟎𝟎.

6. Barry’s mountain bike weighs 𝟔𝟔 pounds more than Andy’s. If their bikes weigh 𝟒𝟒𝟒𝟒 pounds altogether, how much does Barry’s bike weigh? Identify the if-then moves in your solution.

If we let 𝒂𝒂 represent the weight in pounds of Andy’s bike, then 𝒂𝒂 + 𝟔𝟔 represents the weight in pounds of Barry’s bike.

𝒂𝒂 + (𝒂𝒂 + 𝟔𝟔) = 𝟒𝟒𝟒𝟒 (𝒂𝒂 + 𝒂𝒂) + 𝟔𝟔 = 𝟒𝟒𝟒𝟒

𝟒𝟒𝒂𝒂 + 𝟔𝟔 = 𝟒𝟒𝟒𝟒 𝟒𝟒𝒂𝒂 + 𝟔𝟔 − 𝟔𝟔 = 𝟒𝟒𝟒𝟒 − 𝟔𝟔

𝟒𝟒𝒂𝒂 + 𝟎𝟎 = 𝟑𝟑𝟔𝟔 𝟒𝟒𝒂𝒂 = 𝟑𝟑𝟔𝟔

𝟏𝟏𝟒𝟒∙ 𝟒𝟒𝒂𝒂 =

𝟏𝟏𝟒𝟒∙ 𝟑𝟑𝟔𝟔

𝟏𝟏 ∙ 𝒂𝒂 = 𝟏𝟏𝟖𝟖 𝒂𝒂 = 𝟏𝟏𝟖𝟖

If 𝟒𝟒𝒂𝒂 + 𝟔𝟔 = 𝟒𝟒𝟒𝟒, then 𝟒𝟒𝒂𝒂 + 𝟔𝟔 − 𝟔𝟔 = 𝟒𝟒𝟒𝟒 − 𝟔𝟔.

If 𝟒𝟒𝒂𝒂 = 𝟑𝟑𝟔𝟔, then 𝟏𝟏𝟒𝟒∙ 𝟒𝟒𝒂𝒂 =

𝟏𝟏𝟒𝟒∙ 𝟑𝟑𝟔𝟔.

Barry's Bike: 𝒂𝒂 + 𝟔𝟔

(𝟏𝟏𝟖𝟖) + 𝟔𝟔 = 𝟒𝟒𝟒𝟒

Barry’s bike weighs 𝟒𝟒𝟒𝟒 pounds.

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7. Trevor and Marissa together have 𝟒𝟒𝟔𝟔 T-shirts to sell. If Marissa has 𝟔𝟔 fewer T-shirts than Trevor, find how many T-shirts Trevor has. Identify the if-then moves in your solution.

Let 𝒕𝒕 represent the number of T-shirts that Trevor has, and let 𝒕𝒕 − 𝟔𝟔 represent the number of T-shirts that Marissa has.

𝒕𝒕 + (𝒕𝒕 − 𝟔𝟔) = 𝟒𝟒𝟔𝟔 (𝒕𝒕+ 𝒕𝒕) + (−𝟔𝟔) = 𝟒𝟒𝟔𝟔

𝟒𝟒𝒕𝒕 + (−𝟔𝟔) = 𝟒𝟒𝟔𝟔 𝟒𝟒𝒕𝒕+ (−𝟔𝟔) + 𝟔𝟔 = 𝟒𝟒𝟔𝟔 + 𝟔𝟔

𝟒𝟒𝒕𝒕 + 𝟎𝟎 = 𝟑𝟑𝟒𝟒 𝟒𝟒𝒕𝒕 = 𝟑𝟑𝟒𝟒

𝟏𝟏𝟒𝟒∙ 𝟒𝟒𝒕𝒕 =

𝟏𝟏𝟒𝟒∙ 𝟑𝟑𝟒𝟒

𝟏𝟏 ∙ 𝒕𝒕 = 𝟏𝟏𝟔𝟔 𝒕𝒕 = 𝟏𝟏𝟔𝟔

If-then move: Addition property of equality

If-then move: Multiplication property of equality

Trevor has 𝟏𝟏𝟔𝟔 T-shirts to sell, and Marissa has 𝟏𝟏𝟎𝟎 T-shirts to sell.

8. A number is 𝟏𝟏𝟐𝟐

of another number. The difference of the numbers is 𝟏𝟏𝟖𝟖. (Assume that you are subtracting the

smaller number from the larger number.) Find the numbers.

If we let 𝒏𝒏 represent a number, then 𝟏𝟏𝟐𝟐𝒏𝒏 represents the other number.

𝒏𝒏 − �𝟏𝟏𝟐𝟐𝒏𝒏� = 𝟏𝟏𝟖𝟖

𝟐𝟐𝟐𝟐𝒏𝒏 −

𝟏𝟏𝟐𝟐𝒏𝒏 = 𝟏𝟏𝟖𝟖

𝟔𝟔𝟐𝟐𝒏𝒏 = 𝟏𝟏𝟖𝟖

𝟐𝟐𝟔𝟔∙𝟔𝟔𝟐𝟐𝒏𝒏 =

𝟐𝟐𝟔𝟔∙ 𝟏𝟏𝟖𝟖

𝟏𝟏𝒏𝒏 = 𝟐𝟐 ∙ 𝟑𝟑 𝒏𝒏 = 𝟒𝟒𝟏𝟏

The numbers are 𝟒𝟒𝟏𝟏 and 𝟑𝟑.

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9. A number is 𝟔𝟔 greater than 𝟏𝟏𝟒𝟒

another number. If the sum of the numbers is 𝟒𝟒𝟏𝟏, find the numbers.

If we let 𝒏𝒏 represent a number, then 𝟏𝟏𝟒𝟒𝒏𝒏 + 𝟔𝟔 represents the first number.

𝒏𝒏 + �𝟏𝟏𝟒𝟒𝒏𝒏 + 𝟔𝟔� = 𝟒𝟒𝟏𝟏

�𝒏𝒏+𝟏𝟏𝟒𝟒𝒏𝒏� + 𝟔𝟔 = 𝟒𝟒𝟏𝟏

�𝟒𝟒𝟒𝟒𝒏𝒏+

𝟏𝟏𝟒𝟒𝒏𝒏� + 𝟔𝟔 = 𝟒𝟒𝟏𝟏

𝟑𝟑𝟒𝟒𝒏𝒏 + 𝟔𝟔 = 𝟒𝟒𝟏𝟏

𝟑𝟑𝟒𝟒𝒏𝒏 + 𝟔𝟔 − 𝟔𝟔 = 𝟒𝟒𝟏𝟏 − 𝟔𝟔

𝟑𝟑𝟒𝟒𝒏𝒏 + 𝟎𝟎 = 𝟏𝟏𝟓𝟓

𝟑𝟑𝟒𝟒𝒏𝒏 = 𝟏𝟏𝟓𝟓

𝟒𝟒𝟑𝟑∙𝟑𝟑𝟒𝟒𝒏𝒏 =

𝟒𝟒𝟑𝟑∙ 𝟏𝟏𝟓𝟓

𝟏𝟏𝒏𝒏 = 𝟒𝟒 ∙ 𝟓𝟓 𝒏𝒏 = 𝟏𝟏𝟎𝟎

Since the numbers sum to 𝟒𝟒𝟏𝟏, they are 𝟏𝟏𝟎𝟎 and 𝟏𝟏𝟏𝟏.

10. Kevin is currently twice as old as his brother. If Kevin was 𝟖𝟖 years old 𝟒𝟒 years ago, how old is Kevin’s brother now?

If we let 𝒃𝒃 represent Kevin’s brother’s age in years, then Kevin’s age in years is 𝟒𝟒𝒃𝒃.

𝟒𝟒𝒃𝒃 − 𝟒𝟒 = 𝟖𝟖 𝟒𝟒𝒃𝒃 − 𝟒𝟒 + 𝟒𝟒 = 𝟖𝟖 + 𝟒𝟒

𝟒𝟒𝒃𝒃 = 𝟏𝟏𝟎𝟎

�𝟏𝟏𝟒𝟒� (𝟒𝟒𝒃𝒃) = �

𝟏𝟏𝟒𝟒� (𝟏𝟏𝟎𝟎)

𝒃𝒃 = 𝟓𝟓

Kevin’s brother is currently 𝟓𝟓 years old.

11. The sum of two consecutive odd numbers is 𝟏𝟏𝟓𝟓𝟔𝟔. What are the numbers?

If we let 𝒏𝒏 represent one odd number, then 𝒏𝒏 + 𝟒𝟒 represents the next consecutive odd number.

𝒏𝒏 + (𝒏𝒏 + 𝟒𝟒) = 𝟏𝟏𝟓𝟓𝟔𝟔 𝟒𝟒𝒏𝒏 + 𝟒𝟒 − 𝟒𝟒 = 𝟏𝟏𝟓𝟓𝟔𝟔 − 𝟒𝟒

𝟒𝟒𝒏𝒏 = 𝟏𝟏𝟓𝟓𝟒𝟒

�𝟏𝟏𝟒𝟒� (𝟒𝟒𝒏𝒏) = �

𝟏𝟏𝟒𝟒� (𝟏𝟏𝟓𝟓𝟒𝟒)

𝒏𝒏 = 𝟐𝟐𝟐𝟐

The two numbers are 𝟐𝟐𝟐𝟐 and 𝟐𝟐𝟗𝟗.

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12. If 𝒏𝒏 represents an odd integer, write expressions in terms of 𝒏𝒏 that represent the next three consecutive odd integers. If the four consecutive odd integers have a sum of 𝟓𝟓𝟔𝟔, find the numbers.

If we let 𝒏𝒏 represent an odd integer, then 𝒏𝒏 + 𝟒𝟒, 𝒏𝒏 + 𝟒𝟒, and 𝒏𝒏 + 𝟔𝟔 represent the next three consecutive odd integers.

𝒏𝒏 + (𝒏𝒏 + 𝟒𝟒) + (𝒏𝒏+ 𝟒𝟒) + (𝒏𝒏+ 𝟔𝟔) = 𝟓𝟓𝟔𝟔 𝟒𝟒𝒏𝒏 + 𝟏𝟏𝟒𝟒 = 𝟓𝟓𝟔𝟔

𝟒𝟒𝒏𝒏 + 𝟏𝟏𝟒𝟒 − 𝟏𝟏𝟒𝟒 = 𝟓𝟓𝟔𝟔 − 𝟏𝟏𝟒𝟒 𝟒𝟒𝒏𝒏 = 𝟒𝟒𝟒𝟒 𝒏𝒏 = 𝟏𝟏𝟏𝟏

The numbers are 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟑𝟑, 𝟏𝟏𝟓𝟓, and 𝟏𝟏𝟐𝟐.

13. The cost of admission to a history museum is $𝟑𝟑.𝟒𝟒𝟓𝟓 per person over the age of 𝟑𝟑; kids 𝟑𝟑 and under get in for free. If the total cost of admission for the Warrick family, including their two 6-month old twins, is $𝟏𝟏𝟗𝟗.𝟓𝟓𝟎𝟎, find how many family members are over 𝟑𝟑 years old.

If we let 𝒘𝒘 represent the number of Warrick family members, then 𝒘𝒘− 𝟒𝟒 represents the number of family members over the age of 𝟑𝟑 years.

𝟑𝟑.𝟒𝟒𝟓𝟓(𝒘𝒘− 𝟒𝟒) = 𝟏𝟏𝟗𝟗.𝟓𝟓 𝟑𝟑.𝟒𝟒𝟓𝟓𝒘𝒘− 𝟔𝟔.𝟓𝟓 = 𝟏𝟏𝟗𝟗.𝟓𝟓

𝟑𝟑.𝟒𝟒𝟓𝟓𝒘𝒘− 𝟔𝟔.𝟓𝟓 + 𝟔𝟔.𝟓𝟓 = 𝟏𝟏𝟗𝟗.𝟓𝟓 + 𝟔𝟔.𝟓𝟓 𝟑𝟑.𝟒𝟒𝟓𝟓𝒘𝒘 = 𝟒𝟒𝟔𝟔

𝒘𝒘 = 𝟖𝟖 𝒘𝒘− 𝟒𝟒 = 𝟔𝟔

There are 𝟔𝟔 members of the Warrick family over the age of 𝟑𝟑 years.

14. Six times the sum of three consecutive odd integers is −𝟏𝟏𝟖𝟖. Find the integers.

If we let 𝒏𝒏 represent the first odd integer, then 𝒏𝒏 + 𝟒𝟒 and 𝒏𝒏 + 𝟒𝟒 represent the next two consecutive odd integers.

𝟔𝟔�𝒏𝒏 + (𝒏𝒏+ 𝟒𝟒) + (𝒏𝒏 + 𝟒𝟒)� = −𝟏𝟏𝟖𝟖 𝟔𝟔(𝟑𝟑𝒏𝒏+ 𝟔𝟔) = −𝟏𝟏𝟖𝟖 𝟏𝟏𝟖𝟖𝒏𝒏 + 𝟑𝟑𝟔𝟔 = −𝟏𝟏𝟖𝟖

𝟏𝟏𝟖𝟖𝒏𝒏+ 𝟑𝟑𝟔𝟔 − 𝟑𝟑𝟔𝟔 = −𝟏𝟏𝟖𝟖 − 𝟑𝟑𝟔𝟔 𝟏𝟏𝟖𝟖𝒏𝒏 = −𝟓𝟓𝟒𝟒 𝒏𝒏 = −𝟑𝟑

𝒏𝒏+ 𝟒𝟒 = −𝟏𝟏

𝒏𝒏 + 𝟒𝟒 = 𝟏𝟏

The integers are −𝟑𝟑, −𝟏𝟏, and 𝟏𝟏.

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15. I am thinking of a number. If you multiply my number by 𝟒𝟒, add −𝟒𝟒 to the product, and then take 𝟏𝟏𝟑𝟑

of the sum, the

result is −𝟔𝟔. Find my number.

Let 𝒏𝒏 represent the given number.

𝟏𝟏𝟑𝟑 �𝟒𝟒𝒏𝒏 + (−𝟒𝟒)� = −𝟔𝟔

𝟒𝟒𝟑𝟑𝒏𝒏 −

𝟒𝟒𝟑𝟑

= −𝟔𝟔

𝟒𝟒𝟑𝟑𝒏𝒏−

𝟒𝟒𝟑𝟑

+𝟒𝟒𝟑𝟑

= −𝟔𝟔 +𝟒𝟒𝟑𝟑

𝟒𝟒𝟑𝟑𝒏𝒏 =

−𝟏𝟏𝟒𝟒𝟑𝟑

𝒏𝒏 = −𝟑𝟑𝟏𝟏𝟒𝟒

16. A vending machine has twice as many quarters in it as dollar bills. If the quarters and dollar bills have a combined value of $𝟗𝟗𝟔𝟔.𝟎𝟎𝟎𝟎, how many quarters are in the machine?

If we let 𝒅𝒅 represent the number of dollar bills in the machine, then 𝟒𝟒𝒅𝒅 represents the number of quarters in the machine.

𝟒𝟒𝒅𝒅 ∙ �𝟏𝟏𝟒𝟒� + 𝟏𝟏𝒅𝒅 ∙ (𝟏𝟏) = 𝟗𝟗𝟔𝟔

𝟏𝟏𝟒𝟒𝒅𝒅+ 𝟏𝟏𝒅𝒅 = 𝟗𝟗𝟔𝟔

𝟏𝟏𝟏𝟏𝟒𝟒𝒅𝒅 = 𝟗𝟗𝟔𝟔

𝟑𝟑𝟒𝟒𝒅𝒅 = 𝟗𝟗𝟔𝟔

𝟒𝟒𝟑𝟑�𝟑𝟑𝟒𝟒𝒅𝒅� =

𝟒𝟒𝟑𝟑

(𝟗𝟗𝟔𝟔)

𝒅𝒅 = 𝟔𝟔𝟒𝟒 𝟒𝟒𝒅𝒅 = 𝟏𝟏𝟒𝟒𝟖𝟖

There are 𝟏𝟏𝟒𝟒𝟖𝟖 quarters in the machine.

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Student Outcomes

Students understand and use the addition, subtraction, multiplication, division, and substitution properties of equality to solve word problems leading to equations of the form 𝑝𝑝𝑝𝑝 + 𝑞𝑞 = 𝑟𝑟 and 𝑝𝑝(𝑝𝑝 + 𝑞𝑞) = 𝑟𝑟, where 𝑝𝑝, 𝑞𝑞, and 𝑟𝑟 are specific rational numbers.

Students understand that any equation can be rewritten as an equivalent equation with expressions that involve only integer coefficients by multiplying both sides by the correct number.

Lesson Notes This lesson is a continuation from Lesson 8. Students examine and interpret the structure between 𝑝𝑝𝑝𝑝 + 𝑞𝑞 = 𝑟𝑟 and 𝑝𝑝(𝑝𝑝 + 𝑞𝑞) = 𝑟𝑟. Students continue to write equations from word problems including distance and age problems. Also, students play a game during this lesson, which requires students to solve 1–2 problems and then arrange the answers in correct numerical order. This game can be played many different times as long as students receive different problems each time.

Classwork


Have students work in small groups to write and solve an equation for each problem, followed by a whole group discussion.

Opening Exercise

Heather practices soccer and piano. Each day she practices piano for 𝟐𝟐 hours. After 𝟓𝟓 days, she practiced both piano and soccer for a total of 𝟐𝟐𝟐𝟐 hours. Assuming that she practiced soccer the same amount of time each day, how many hours per day, 𝒉𝒉, did Heather practice soccer?

𝒉𝒉: hours per day that soccer was practiced

𝟓𝟓(𝒉𝒉 + 𝟐𝟐) = 𝟐𝟐𝟐𝟐 𝟓𝟓𝒉𝒉+ 𝟏𝟏𝟐𝟐 = 𝟐𝟐𝟐𝟐

𝟓𝟓𝒉𝒉 + 𝟏𝟏𝟐𝟐 − 𝟏𝟏𝟐𝟐 = 𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟐𝟐 𝟓𝟓𝒉𝒉 = 𝟏𝟏𝟐𝟐

�𝟏𝟏𝟓𝟓� (𝟓𝟓𝒉𝒉) = �

𝟏𝟏𝟓𝟓� (𝟏𝟏𝟐𝟐)

𝒉𝒉 = 𝟐𝟐 Heather practiced soccer for 𝟐𝟐 hours each day.

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Over 𝟓𝟓 days, Jake practices piano for a total of 𝟐𝟐 hours. Jake practices soccer for the same amount of time each day. If he practiced piano and soccer for a total of 𝟐𝟐𝟐𝟐 hours, how many hours, 𝒉𝒉, per day did Jake practice soccer?

𝒉𝒉: hours per day that soccer was practiced

𝟓𝟓𝒉𝒉+ 𝟐𝟐 = 𝟐𝟐𝟐𝟐 𝟓𝟓𝒉𝒉 + 𝟐𝟐 − 𝟐𝟐 = 𝟐𝟐𝟐𝟐 − 𝟐𝟐

𝟓𝟓𝒉𝒉 = 𝟏𝟏𝟏𝟏

�𝟏𝟏𝟓𝟓� (𝟓𝟓𝒉𝒉) = (𝟏𝟏𝟏𝟏) �

𝟏𝟏𝟓𝟓�

𝒉𝒉 = 𝟑𝟑.𝟔𝟔 Jake practiced soccer 𝟑𝟑.𝟔𝟔 hours each day.

Examine both equations. How are they similar, and how are they different?

Both equations have the same numbers and deal with the same word problem. They are different in the set-up of the equations. The first problem includes parentheses where the second does not. This is because, in the first problem, both soccer and piano were being practiced every day, so the total for each day had to be multiplied by the total number of days, five. Whereas in the second problem, only soccer was being practiced every day, and piano was only practiced a total of two hours for that time frame. Therefore, only the number of hours of soccer practice had to be multiplied by five, and not the piano time.

Do the different structures of the equations affect the answer? Explain why or why not. Yes, the first problem requires students to use the distributive property, so the number of hours of

soccer and piano practice are included every day. Using the distributive property changes the 2 in the equation to 10, which is the total hours of piano practice over the entire 5 days. An if-then move of dividing both sides by 5 first could have also been used to solve the problem. The second equation does not use parentheses since piano is not practiced every day. Therefore, the 5 days are only multiplied by the number of hours of soccer practice and not the piano time. This changes the end result.

Which if-then moves were used in solving the equations?

In the first equation, students may have used division of a number on both sides, subtracting a number on both sides, and multiplying a number on both sides. If the student distributed first, then only the if-then moves of subtracting a number on both sides and multiplying a nonzero number on both sides were used.

In the second equation, the if-then moves of subtracting a number on both sides and multiplying a nonzero number on both sides were used.

Interpret what 3.6 hours means in hours and minutes? Describe how to determine this.

The solution 3.6 hours means 3 hours 36 minutes. Since there are 60 minutes in an hour and 0.6 is part of an hour, multiply 0.6 by 60 to get the part of the hour that 0.6 represents.

MP.2

MP.7

MP.2

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Lead students through the following problem. Example 1

Fred and Sam are a team in the local 𝟏𝟏𝟑𝟑𝟏𝟏.𝟐𝟐 mile bike-run-athon. Fred will compete in the bike race, and Sam will compete in the run. Fred bikes at an average speed of 𝟏𝟏 miles per hour and Sam runs at an average speed of 𝟒𝟒 miles per hour. The bike race begins at 6:00 a.m., followed by the run. Sam predicts he will finish the run at 2:33 a.m. the next morning.

a. How many hours will it take them to complete the entire bike-run-athon?

From 6:00 a.m. to 2:00 a.m. the following day is 𝟐𝟐𝟐𝟐 hours.

𝟑𝟑𝟑𝟑 minutes in hours is 𝟑𝟑𝟑𝟑𝟔𝟔𝟐𝟐

=𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

= 𝟐𝟐. 𝟓𝟓𝟓𝟓, or 𝟐𝟐.𝟓𝟓𝟓𝟓 hours.

Therefore, the total time it will take to complete the entire bike-run-athon is 𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 hours.

b. If 𝒕𝒕 is how long it takes Fred to complete the bike race, in hours, write an expression to find Fred’s total distance.

𝒅𝒅 = 𝒓𝒓𝒕𝒕 𝒅𝒅 = 𝟏𝟏𝒕𝒕

The expression of Fred’s total distance is 𝟏𝟏𝒕𝒕.

c. Write an expression, in terms of 𝒕𝒕 to express Sam’s time.

Since 𝒕𝒕 is Fred’s time and 𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 is the total time, then Sam’s time would be the difference between the total time and Fred’s time. The expression would be 𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝒕𝒕.

d. Write an expression, in terms of 𝒕𝒕, that represents Sam’s total distance.

𝒅𝒅 = 𝒓𝒓𝒕𝒕 𝒅𝒅 = 𝟒𝟒(𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝒕𝒕)

The expressions 𝟒𝟒(𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝒕𝒕) or 𝟏𝟏𝟐𝟐.𝟐𝟐 − 𝟒𝟒𝒕𝒕 is Sam’s total distance.

e. Write and solve an equation using the total distance both Fred and Sam will travel.

𝟏𝟏𝒕𝒕 + 𝟒𝟒(𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝒕𝒕) = 𝟏𝟏𝟑𝟑𝟏𝟏.𝟐𝟐 𝟏𝟏𝒕𝒕+ 𝟏𝟏𝟐𝟐.𝟐𝟐 − 𝟒𝟒𝒕𝒕 = 𝟏𝟏𝟑𝟑𝟏𝟏.𝟐𝟐 𝟏𝟏𝒕𝒕 − 𝟒𝟒𝒕𝒕 + 𝟏𝟏𝟐𝟐.𝟐𝟐 = 𝟏𝟏𝟑𝟑𝟏𝟏.𝟐𝟐

𝟒𝟒𝒕𝒕 + 𝟏𝟏𝟐𝟐.𝟐𝟐 = 𝟏𝟏𝟑𝟑𝟏𝟏.𝟐𝟐 𝟒𝟒𝒕𝒕 + 𝟏𝟏𝟐𝟐.𝟐𝟐 − 𝟏𝟏𝟐𝟐.𝟐𝟐 = 𝟏𝟏𝟑𝟑𝟏𝟏.𝟐𝟐 − 𝟏𝟏𝟐𝟐.𝟐𝟐

𝟒𝟒𝒕𝒕 + 𝟐𝟐 = 𝟓𝟓𝟔𝟔

�𝟏𝟏𝟒𝟒� (𝟒𝟒𝒕𝒕) = �

𝟏𝟏𝟒𝟒� (𝟓𝟓𝟔𝟔)

𝒕𝒕 = 𝟏𝟏𝟒𝟒

Fred’s time: 𝟏𝟏𝟒𝟒 hours

Sam’s time: 𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝒕𝒕 = 𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝟏𝟏𝟒𝟒 = 𝟔𝟔.𝟓𝟓𝟓𝟓

𝟔𝟔.𝟓𝟓𝟓𝟓 hours

Scaffolding:

Refer to a clock when determining the total amount of time.

Teachers may need to review the formula 𝑑𝑑 = 𝑟𝑟𝑟𝑟 from Grade 6 and Module 1.

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f. How far will Fred bike, and how much time will it take him to complete his leg of the race?

𝟏𝟏(𝟏𝟏𝟒𝟒) = 𝟏𝟏𝟏𝟏𝟐𝟐

Fred will bike 𝟏𝟏𝟏𝟏𝟐𝟐 miles and will complete the bike race in 𝟏𝟏𝟒𝟒 hours.

g. How far will Sam run, and how much time will it take him to complete his leg of the race?

𝟒𝟒(𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝒕𝒕)

𝟒𝟒(𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝟏𝟏𝟒𝟒)

𝟒𝟒(𝟔𝟔.𝟓𝟓𝟓𝟓)

𝟐𝟐𝟔𝟔.𝟐𝟐

Sam will run 𝟐𝟐𝟔𝟔.𝟐𝟐 miles, and it will take him 𝟔𝟔.𝟓𝟓𝟓𝟓 hours.


Why isn’t the total time from 6:00 a.m. to 2:33 a.m. written as 20.33 hours?

Time is based on 60 minutes. If the time in minutes just became the decimal, then time would have to be out of 100 because 20.33 represents 20 and 33 hundredths.

To help determine the expression for Sam’s time, work through the following chart. (This will lead to subtracting Fred’s time from the total time.)

Total Time (hours) Fred’s Time (hours) Sam’s Time (hours)

𝟏𝟏𝟐𝟐 𝟔𝟔 𝟏𝟏𝟐𝟐 − 𝟔𝟔 = 𝟒𝟒 𝟏𝟏𝟓𝟓 𝟏𝟏𝟐𝟐 𝟏𝟏𝟓𝟓 − 𝟏𝟏𝟐𝟐 = 𝟑𝟑 𝟐𝟐𝟐𝟐 𝟏𝟏 𝟐𝟐𝟐𝟐 − 𝟏𝟏 = 𝟏𝟏𝟐𝟐

𝟏𝟏𝟏𝟏.𝟑𝟑𝟓𝟓 𝟏𝟏 𝟏𝟏𝟏𝟏.𝟑𝟑𝟓𝟓 − 𝟏𝟏 = 𝟏𝟏𝟐𝟐.𝟑𝟑𝟓𝟓 𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 𝒕𝒕 𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝒕𝒕

How do you find the distance traveled?

Multiply the rate of speed by the amount of time.

Model how to organize the problem in a distance, rate, and time chart.

Rate (mph) Time (hours) Distance (miles)

Fred 𝟏𝟏 𝒕𝒕 𝟏𝟏𝒕𝒕

Sam 𝟒𝟒 𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝒕𝒕 𝟒𝟒(𝟐𝟐𝟐𝟐.𝟓𝟓𝟓𝟓 − 𝒕𝒕) 𝟏𝟏𝟐𝟐.𝟐𝟐 − 𝟒𝟒𝒕𝒕

Explain how to write the equation to have only integers and no decimals. Write the equation.

Since the decimal terminates in the tenths place, if we multiply every term by 10, the equation would result with only integer coefficients. The equation would be 40𝑟𝑟 + 822 = 1382.

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Example 2

Shelby is seven times as old as Bonnie. If in 𝟓𝟓 years, the sum of Bonnie’s and Shelby’s ages is 𝟗𝟗𝟏𝟏, find Bonnie’s present age. Use an algebraic approach.

Present Age (in years) Future Age (in years) Bonnie 𝒙𝒙 𝒙𝒙 + 𝟓𝟓 Shelby 𝟕𝟕𝒙𝒙 𝟕𝟕𝒙𝒙 + 𝟓𝟓

𝒙𝒙 + 𝟓𝟓 + 𝟕𝟕𝒙𝒙 + 𝟓𝟓 = 𝟗𝟗𝟏𝟏 𝟏𝟏𝒙𝒙 + 𝟏𝟏𝟐𝟐 = 𝟗𝟗𝟏𝟏

𝟏𝟏𝒙𝒙 + 𝟏𝟏𝟐𝟐 − 𝟏𝟏𝟐𝟐 = 𝟗𝟗𝟏𝟏 − 𝟏𝟏𝟐𝟐 𝟏𝟏𝒙𝒙 = 𝟏𝟏𝟏𝟏

�𝟏𝟏𝟏𝟏� (𝟏𝟏𝒙𝒙) = �

𝟏𝟏𝟏𝟏� (𝟏𝟏𝟏𝟏)

𝒙𝒙 = 𝟏𝟏𝟏𝟏 Bonnie’s present age is 𝟏𝟏𝟏𝟏 years old.

The first step we must take is to write expressions that represent the present ages of both Bonnie and Shelby. The second step is to write expressions for future time or past time, using the present age expressions. How would the expression change if the time were in the past and not in the future?

If the time were in the past, then the expression would be the difference between the present age and the amount of time in the past.

Game (10 minutes)

The purpose of this game is for students to continue to practice solving linear equations when given in a contextual form. Divide students into 3 groups. There are 25 problems total, so if there are more than 25 students in the class, assign the extra students as the checkers of student work. Each group receives a puzzle (found at the end of the lesson). Depending on the size of the class, some students may receive only one card, while others may have multiple cards. Direct students to complete the problem(s) they receive. Each problem has a letter to the right. Students are to write and solve an equation unless other directions are stated. Once students get an answer, they are to locate the numerical answer under the blank and put the corresponding letter in the blank. When all problems are completed correctly, the letters in the blanks answer the riddle. Encourage students to check each other’s work. This game can be replayed as many times as desired provided students receive different problems from a different set of cards. A variation to this game can be for students to arrange the answers in numerical order from least to greatest and/or greatest to least instead of the riddle or in addition to the riddle.

Closing (2 minutes)

How can an equation be written with only integer coefficients and constant terms?

How are the addition, subtraction, multiplication, division, and substitution properties of equality used to solve algebraic equations?


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Name ___________________________________________________ Date____________________


Exit Ticket 1. Brand A scooter has a top speed that goes 2 miles per hour faster than Brand B. If after 3 hours, Brand A scooter

traveled 24 miles at its top speed, at what rate did Brand B scooter travel at its top speed if it traveled the same distance? Write an equation to determine the solution. Identify the if-then moves used in your solution.

2. At each scooter’s top speed, Brand A scooter goes 2 miles per hour faster than Brand B. If after traveling at its top speed for 3 hours, Brand A scooter traveled 40.2 miles, at what rate did Brand B scooter travel if it traveled the same distance as Brand A? Write an equation to determine the solution and then write an equivalent equation using only integers.

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1. Brand A scooter has a top speed that goes 𝟐𝟐 miles per hour faster than Brand B. If after 𝟑𝟑 hours, Brand A scooter traveled 𝟐𝟐𝟒𝟒 miles at its top speed, at what rate did Brand B scooter travel at its top speed if it traveled the same distance? Write an equation to determine the solution. Identify the if-then moves used in your solution.

𝒙𝒙: speed, in mph, of Brand B scooter

𝒙𝒙 + 𝟐𝟐: speed, in mph, of Brand A scooter

𝒅𝒅 = 𝒓𝒓𝒕𝒕 𝟐𝟐𝟒𝟒 = (𝒙𝒙 + 𝟐𝟐)(𝟑𝟑) 𝟐𝟐𝟒𝟒 = 𝟑𝟑(𝒙𝒙+ 𝟐𝟐)

Possible solution 1: Possible solution 2:

If-then Moves: Divide both sides by 𝟑𝟑. If-then Moves: Subtract 𝟔𝟔 from both sides.

Subtract 𝟐𝟐 from both sides. Multiply both sides by 𝟏𝟏𝟑𝟑.

2. At each scooter’s top speed, Brand A scooter goes 𝟐𝟐 miles per hour faster than Brand B. If after traveling at its top speed for 3 hours, Brand A scooter traveled 𝟒𝟒𝟐𝟐.𝟐𝟐 miles, at what rate did Brand B scooter travel if it traveled the same distance as Brand A? Write an equation to determine the solution and then write an equivalent equation using only integers.

𝒙𝒙: speed, in mph, of Brand B scooter

𝒙𝒙 + 𝟐𝟐: speed, in mph, of Brand A scooter

𝒅𝒅 = 𝒓𝒓𝒕𝒕 𝟒𝟒𝟐𝟐.𝟐𝟐 = (𝒙𝒙 + 𝟐𝟐)(𝟑𝟑) 𝟒𝟒𝟐𝟐.𝟐𝟐 = 𝟑𝟑(𝒙𝒙 + 𝟐𝟐)

Possible solution 1: Possible solution 2:

Brand B's scooter travels at 𝟏𝟏𝟏𝟏.𝟒𝟒 miles per hour.

𝟒𝟒𝟐𝟐.𝟐𝟐 = 𝟑𝟑(𝒙𝒙+ 𝟐𝟐) 𝟏𝟏𝟑𝟑.𝟒𝟒 = 𝒙𝒙 + 𝟐𝟐 𝟏𝟏𝟑𝟑𝟒𝟒 = 𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟐𝟐𝟐𝟐

𝟏𝟏𝟑𝟑𝟒𝟒 − 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟐𝟐𝟐𝟐 − 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟒𝟒 = 𝟏𝟏𝟐𝟐𝒙𝒙

�𝟏𝟏𝟏𝟏𝟐𝟐� (𝟏𝟏𝟏𝟏𝟒𝟒) = �

𝟏𝟏𝟏𝟏𝟐𝟐� (𝟏𝟏𝟐𝟐𝒙𝒙)

𝟏𝟏𝟏𝟏.𝟒𝟒 = 𝒙𝒙

𝟒𝟒𝟐𝟐.𝟐𝟐 = 𝟑𝟑(𝒙𝒙+ 𝟐𝟐) 𝟒𝟒𝟐𝟐.𝟐𝟐 = 𝟑𝟑𝒙𝒙 + 𝟔𝟔 𝟒𝟒𝟐𝟐𝟐𝟐 = 𝟑𝟑𝟐𝟐𝒙𝒙 + 𝟔𝟔𝟐𝟐

𝟒𝟒𝟐𝟐𝟐𝟐 − 𝟔𝟔𝟐𝟐 = 𝟑𝟑𝟐𝟐𝒙𝒙 + 𝟔𝟔𝟐𝟐 − 𝟔𝟔𝟐𝟐 𝟑𝟑𝟒𝟒𝟐𝟐 = 𝟑𝟑𝟐𝟐𝒙𝒙

�𝟏𝟏𝟑𝟑𝟐𝟐� (𝟑𝟑𝟒𝟒𝟐𝟐) = �

𝟏𝟏𝟑𝟑𝟐𝟐� (𝟑𝟑𝟐𝟐𝒙𝒙)

𝟏𝟏𝟏𝟏.𝟒𝟒 = 𝒙𝒙

𝟐𝟐𝟒𝟒 = 𝟑𝟑(𝒙𝒙 + 𝟐𝟐) 𝟏𝟏 = 𝒙𝒙 + 𝟐𝟐

𝟏𝟏 − 𝟐𝟐 = 𝒙𝒙 + 𝟐𝟐 − 𝟐𝟐 𝟔𝟔 = 𝒙𝒙

𝟐𝟐𝟒𝟒 = 𝟑𝟑(𝒙𝒙 + 𝟐𝟐) 𝟐𝟐𝟒𝟒 = 𝟑𝟑𝒙𝒙 + 𝟔𝟔

𝟐𝟐𝟒𝟒 − 𝟔𝟔 = 𝟑𝟑𝒙𝒙 + 𝟔𝟔 − 𝟔𝟔 𝟏𝟏𝟏𝟏 = 𝟑𝟑𝒙𝒙 + 𝟐𝟐

�𝟏𝟏𝟑𝟑� (𝟏𝟏𝟏𝟏) = �

𝟏𝟏𝟑𝟑� (𝟑𝟑𝒙𝒙)

𝟔𝟔 = 𝒙𝒙

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7•3 Lesson 9



1. A company buys a digital scanner for $𝟏𝟏𝟐𝟐,𝟐𝟐𝟐𝟐𝟐𝟐. The value of the scanner is 𝟏𝟏𝟐𝟐,𝟐𝟐𝟐𝟐𝟐𝟐�𝟏𝟏 − 𝒏𝒏𝟓𝟓� after 𝒏𝒏 years. The

company has budgeted to replace the scanner when the trade-in value is $𝟐𝟐,𝟒𝟒𝟐𝟐𝟐𝟐. After how many years should the company plan to replace the machine in order to receive this trade-in value?

𝟏𝟏𝟐𝟐,𝟐𝟐𝟐𝟐𝟐𝟐�𝟏𝟏 −𝒏𝒏𝟓𝟓� = 𝟐𝟐,𝟒𝟒𝟐𝟐𝟐𝟐

𝟏𝟏𝟐𝟐,𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟐𝟐,𝟒𝟒𝟐𝟐𝟐𝟐𝒏𝒏 = 𝟐𝟐,𝟒𝟒𝟐𝟐𝟐𝟐 −𝟐𝟐,𝟒𝟒𝟐𝟐𝟐𝟐𝒏𝒏+ 𝟏𝟏𝟐𝟐,𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟐𝟐,𝟐𝟐𝟐𝟐𝟐𝟐 = 𝟐𝟐,𝟒𝟒𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟐𝟐,𝟐𝟐𝟐𝟐𝟐𝟐

−𝟐𝟐,𝟒𝟒𝟐𝟐𝟐𝟐𝒏𝒏 = −𝟗𝟗,𝟔𝟔𝟐𝟐𝟐𝟐 𝒏𝒏 = 𝟒𝟒

They will replace the scanner after 𝟒𝟒 years.

2. Michael is 𝟏𝟏𝟕𝟕 years older than John. In 𝟒𝟒 years, the sum of their ages will be 𝟒𝟒𝟗𝟗. Find Michael’s present age.

𝒙𝒙 represents Michael’s age now in years.

Now 𝟒𝟒 years later Michael 𝒙𝒙 𝒙𝒙 + 𝟒𝟒

John 𝒙𝒙 − 𝟏𝟏𝟕𝟕 (𝒙𝒙 − 𝟏𝟏𝟕𝟕) + 𝟒𝟒

𝒙𝒙 + 𝟒𝟒 + 𝒙𝒙 − 𝟏𝟏𝟕𝟕+ 𝟒𝟒 = 𝟒𝟒𝟗𝟗 𝒙𝒙 + 𝟒𝟒+ 𝒙𝒙 − 𝟏𝟏𝟑𝟑 = 𝟒𝟒𝟗𝟗

𝟐𝟐𝒙𝒙 − 𝟗𝟗 = 𝟒𝟒𝟗𝟗 𝟐𝟐𝒙𝒙 − 𝟗𝟗 + 𝟗𝟗 = 𝟒𝟒𝟗𝟗+ 𝟗𝟗

𝟐𝟐𝒙𝒙 = 𝟓𝟓𝟏𝟏

�𝟏𝟏𝟐𝟐� (𝟐𝟐𝒙𝒙) = �

𝟏𝟏𝟐𝟐� (𝟓𝟓𝟏𝟏)

𝒙𝒙 = 𝟐𝟐𝟗𝟗

Michael’s present age is 𝟐𝟐𝟗𝟗 years old.

3. Brady rode his bike 𝟕𝟕𝟐𝟐 miles in 𝟒𝟒 hours. He rode at an average speed of 𝟏𝟏𝟕𝟕 mph for 𝒕𝒕 hours and at an average rate of speed of 𝟐𝟐𝟐𝟐 mph for the rest of the time. How long did Brady ride at the slower speed? Use the variable 𝒕𝒕 to represent the time, in hours, Brady rode at 𝟏𝟏𝟕𝟕 mph.

Rate

(mph) Time

(hours) Distance (miles)

Brady speed 1 𝟏𝟏𝟕𝟕 𝒕𝒕 𝟏𝟏𝟕𝟕𝒕𝒕

Brady speed 2 𝟐𝟐𝟐𝟐 𝟒𝟒 − 𝒕𝒕 𝟐𝟐𝟐𝟐(𝟒𝟒 − 𝒕𝒕)

The total distance he rode: 𝟏𝟏𝟕𝟕𝒕𝒕 + 𝟐𝟐𝟐𝟐(𝟒𝟒 − 𝒕𝒕)

The total distance equals 𝟕𝟕𝟐𝟐 miles:

𝟏𝟏𝟕𝟕𝒕𝒕 + 𝟐𝟐𝟐𝟐(𝟒𝟒 − 𝒕𝒕) = 𝟕𝟕𝟐𝟐 𝟏𝟏𝟕𝟕𝒕𝒕+ 𝟏𝟏𝟏𝟏 − 𝟐𝟐𝟐𝟐𝒕𝒕 = 𝟕𝟕𝟐𝟐

−𝟓𝟓𝒕𝒕+ 𝟏𝟏𝟏𝟏 = 𝟕𝟕𝟐𝟐 −𝟓𝟓𝒕𝒕+ 𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏 = 𝟕𝟕𝟐𝟐 − 𝟏𝟏𝟏𝟏

−𝟓𝟓𝒕𝒕 = −𝟏𝟏𝟏𝟏 𝒕𝒕 = 𝟑𝟑.𝟔𝟔

Brady rode at 𝟏𝟏𝟕𝟕 mph for 𝟑𝟑.𝟔𝟔 hours.

Total distance

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4. Caitlan went to the store to buy school clothes. She had a store credit from a previous return in the amount of $𝟑𝟑𝟗𝟗.𝟓𝟓𝟏𝟏. If she bought 𝟒𝟒 of the same style shirt in different colors and spent a total of $𝟓𝟓𝟐𝟐.𝟐𝟐𝟐𝟐 after the store credit was taken off her total, what was the price of each shirt she bought? Write and solve an equation with integer coefficients.

𝒕𝒕: the price of one shirt

𝟒𝟒𝒕𝒕 − 𝟑𝟑𝟗𝟗.𝟓𝟓𝟏𝟏 = 𝟓𝟓𝟐𝟐.𝟐𝟐𝟐𝟐 𝟒𝟒𝒕𝒕 − 𝟑𝟑𝟗𝟗.𝟓𝟓𝟏𝟏+ 𝟑𝟑𝟗𝟗.𝟓𝟓𝟏𝟏 = 𝟓𝟓𝟐𝟐.𝟐𝟐𝟐𝟐+ 𝟑𝟑𝟗𝟗.𝟓𝟓𝟏𝟏

𝟒𝟒𝒕𝒕 + 𝟐𝟐 = 𝟗𝟗𝟏𝟏.𝟏𝟏𝟐𝟐

�𝟏𝟏𝟒𝟒� (𝟒𝟒𝒕𝒕) = �

𝟏𝟏𝟒𝟒� (𝟗𝟗𝟏𝟏.𝟏𝟏𝟐𝟐)

𝒕𝒕 = 𝟐𝟐𝟐𝟐.𝟗𝟗𝟓𝟓

The price of one shirt was $𝟐𝟐𝟐𝟐.𝟗𝟗𝟓𝟓.

5. A young boy is growing at a rate of 𝟑𝟑.𝟓𝟓 𝐜𝐜𝐜𝐜 per month. He is currently 𝟗𝟗𝟐𝟐 𝐜𝐜𝐜𝐜 tall. At that rate, in how many months will the boy grow to a height of 𝟏𝟏𝟑𝟑𝟐𝟐 𝐜𝐜𝐜𝐜?

Let 𝒎𝒎 represent the number of months.

𝟑𝟑.𝟓𝟓𝒎𝒎+ 𝟗𝟗𝟐𝟐 = 𝟏𝟏𝟑𝟑𝟐𝟐 𝟑𝟑.𝟓𝟓𝒎𝒎 + 𝟗𝟗𝟐𝟐 − 𝟗𝟗𝟐𝟐 = 𝟏𝟏𝟑𝟑𝟐𝟐 − 𝟗𝟗𝟐𝟐

𝟑𝟑.𝟓𝟓𝒎𝒎 = 𝟒𝟒𝟐𝟐

�𝟏𝟏𝟑𝟑.𝟓𝟓

� (𝟑𝟑.𝟓𝟓𝒎𝒎) = �𝟏𝟏𝟑𝟑.𝟓𝟓

� (𝟒𝟒𝟐𝟐)

𝒎𝒎 = 𝟏𝟏𝟐𝟐

The boy will grow to be 𝟏𝟏𝟑𝟑𝟐𝟐 𝐜𝐜𝐜𝐜 tall 𝟏𝟏𝟐𝟐 months from now.

6. The sum of a number, 𝟏𝟏𝟔𝟔

of that number, 𝟐𝟐𝟏𝟏𝟐𝟐 of that number, and 𝟕𝟕 is 1𝟐𝟐 𝟏𝟏𝟐𝟐

. Find the number.

Let 𝒏𝒏 represent the given number.

𝒏𝒏 +𝟏𝟏𝟔𝟔𝒏𝒏 + �𝟐𝟐

𝟏𝟏𝟐𝟐�𝒏𝒏 + 𝟕𝟕 = 𝟏𝟏𝟐𝟐

𝟏𝟏𝟐𝟐

𝒏𝒏�𝟏𝟏+𝟏𝟏𝟔𝟔

+𝟓𝟓𝟐𝟐� + 𝟕𝟕 = 𝟏𝟏𝟐𝟐

𝟏𝟏𝟐𝟐

𝒏𝒏�𝟔𝟔𝟔𝟔

+𝟏𝟏𝟔𝟔

+𝟏𝟏𝟓𝟓𝟔𝟔�+ 𝟕𝟕 = 𝟏𝟏𝟐𝟐

𝟏𝟏𝟐𝟐

𝒏𝒏 �𝟐𝟐𝟐𝟐𝟔𝟔�+ 𝟕𝟕 = 𝟏𝟏𝟐𝟐

𝟏𝟏𝟐𝟐

𝟏𝟏𝟏𝟏𝟑𝟑𝒏𝒏 + 𝟕𝟕 − 𝟕𝟕 = 𝟏𝟏𝟐𝟐

𝟏𝟏𝟐𝟐− 𝟕𝟕

𝟏𝟏𝟏𝟏𝟑𝟑𝒏𝒏 + 𝟐𝟐 = 𝟓𝟓

𝟏𝟏𝟐𝟐

𝟏𝟏𝟏𝟏𝟑𝟑𝒏𝒏 = 𝟓𝟓

𝟏𝟏𝟐𝟐

𝟑𝟑𝟏𝟏𝟏𝟏

∙𝟏𝟏𝟏𝟏𝟑𝟑𝒏𝒏 =

𝟑𝟑𝟏𝟏𝟏𝟏

∙𝟏𝟏𝟏𝟏𝟐𝟐

𝟏𝟏𝒏𝒏 =𝟑𝟑𝟐𝟐

𝒏𝒏 = 𝟏𝟏𝟏𝟏𝟐𝟐

The number is 𝟏𝟏𝟏𝟏𝟐𝟐.

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7. The sum of two numbers is 𝟑𝟑𝟑𝟑 and their difference is 𝟐𝟐. Find the numbers.

Let 𝒙𝒙 represent the first number, then 𝟑𝟑𝟑𝟑 − 𝒙𝒙 represents the other number since their sum is 𝟑𝟑𝟑𝟑.

𝒙𝒙 − (𝟑𝟑𝟑𝟑 − 𝒙𝒙) = 𝟐𝟐 𝒙𝒙 + �−(𝟑𝟑𝟑𝟑 − 𝒙𝒙)� = 𝟐𝟐 𝒙𝒙 + (−𝟑𝟑𝟑𝟑) + 𝒙𝒙 = 𝟐𝟐 𝟐𝟐𝒙𝒙 + (−𝟑𝟑𝟑𝟑) = 𝟐𝟐

𝟐𝟐𝒙𝒙 + (−𝟑𝟑𝟑𝟑) + 𝟑𝟑𝟑𝟑 = 𝟐𝟐+ 𝟑𝟑𝟑𝟑 𝟐𝟐𝒙𝒙 + 𝟐𝟐 = 𝟑𝟑𝟓𝟓

𝟐𝟐𝒙𝒙 = 𝟑𝟑𝟓𝟓 𝟏𝟏𝟐𝟐∙ 𝟐𝟐𝒙𝒙 =

𝟏𝟏𝟐𝟐∙ 𝟑𝟑𝟓𝟓

𝟏𝟏𝒙𝒙 =𝟑𝟑𝟓𝟓𝟐𝟐

𝒙𝒙 = 𝟏𝟏𝟕𝟕𝟏𝟏𝟐𝟐

𝟑𝟑𝟑𝟑 − 𝒙𝒙 = 𝟑𝟑𝟑𝟑 − �𝟏𝟏𝟕𝟕𝟏𝟏𝟐𝟐� = 𝟏𝟏𝟓𝟓

𝟏𝟏𝟐𝟐

�𝟏𝟏𝟕𝟕𝟏𝟏𝟐𝟐

,𝟏𝟏𝟓𝟓𝟏𝟏𝟐𝟐�

8. Aiden refills three token machines in an arcade. He puts twice the number of tokens in machine A as in machine B,

and in machine C, he puts 𝟑𝟑𝟒𝟒

of what he put in machine A. The three machines took a total of 𝟏𝟏𝟏𝟏,𝟑𝟑𝟐𝟐𝟒𝟒 tokens. How

many did each machine take?

Let 𝑨𝑨 represent the number of tokens in machine A. Then 𝟏𝟏𝟐𝟐𝑨𝑨 represents the number of tokens in machine B, and

𝟑𝟑𝟒𝟒𝑨𝑨 represents the number of tokens in machine C.

𝑨𝑨 +𝟏𝟏𝟐𝟐𝑨𝑨 +

𝟑𝟑𝟒𝟒𝑨𝑨 = 𝟏𝟏𝟏𝟏,𝟑𝟑𝟐𝟐𝟒𝟒

𝟗𝟗𝟒𝟒𝑨𝑨 = 𝟏𝟏𝟏𝟏,𝟑𝟑𝟐𝟐𝟒𝟒

𝑨𝑨 = 𝟏𝟏,𝟏𝟏𝟒𝟒𝟒𝟒

Machine A took 𝟏𝟏,𝟏𝟏𝟒𝟒𝟒𝟒 tokens, machine B took 𝟒𝟒,𝟐𝟐𝟕𝟕𝟐𝟐 tokens, and machine C took 𝟔𝟔,𝟏𝟏𝟐𝟐𝟏𝟏 tokens.

9. Paulie ordered 𝟐𝟐𝟓𝟓𝟐𝟐 pens and 𝟐𝟐𝟓𝟓𝟐𝟐 pencils to sell for a theatre club fundraiser. The pens cost 𝟏𝟏𝟏𝟏 cents more than the pencils. If Paulie’s total order costs $𝟒𝟒𝟐𝟐.𝟓𝟓𝟐𝟐, find the cost of each pen and pencil.

Let 𝒍𝒍 represent the cost of a pencil in dollars. Then, the cost of a pen in dollars is 𝒍𝒍 + 𝟐𝟐.𝟏𝟏𝟏𝟏.

𝟐𝟐𝟓𝟓𝟐𝟐(𝒍𝒍 + 𝒍𝒍 + 𝟐𝟐.𝟏𝟏𝟏𝟏) = 𝟒𝟒𝟐𝟐.𝟓𝟓 𝟐𝟐𝟓𝟓𝟐𝟐(𝟐𝟐𝒍𝒍 + 𝟐𝟐.𝟏𝟏𝟏𝟏) = 𝟒𝟒𝟐𝟐.𝟓𝟓

𝟓𝟓𝟐𝟐𝟐𝟐𝒍𝒍 + 𝟐𝟐𝟕𝟕.𝟓𝟓 = 𝟒𝟒𝟐𝟐.𝟓𝟓

𝟓𝟓𝟐𝟐𝟐𝟐𝒍𝒍 + 𝟐𝟐𝟕𝟕.𝟓𝟓+ (−𝟐𝟐𝟕𝟕.𝟓𝟓) = 𝟒𝟒𝟐𝟐.𝟓𝟓 + (−𝟐𝟐𝟕𝟕.𝟓𝟓) 𝟓𝟓𝟐𝟐𝟐𝟐𝒍𝒍 + 𝟐𝟐 = 𝟏𝟏𝟓𝟓

𝟓𝟓𝟐𝟐𝟐𝟐𝒍𝒍 = 𝟏𝟏𝟓𝟓 𝟓𝟓𝟐𝟐𝟐𝟐𝒍𝒍𝟓𝟓𝟐𝟐𝟐𝟐

=𝟏𝟏𝟓𝟓𝟓𝟓𝟐𝟐𝟐𝟐

𝒍𝒍 = 𝟐𝟐.𝟐𝟐𝟑𝟑

A pencil costs $𝟐𝟐.𝟐𝟐𝟑𝟑, and a pen costs $𝟐𝟐.𝟏𝟏𝟒𝟒.

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7•3 Lesson 9


10. A family left their house in two cars at the same time. One car traveled an average of 𝟕𝟕 miles per hour faster than

the other. When the first car arrived at the destination after 𝟓𝟓𝟏𝟏𝟐𝟐 hours of driving, both cars had driven a total of 𝟓𝟓𝟗𝟗𝟗𝟗.𝟓𝟓 miles. If the second car continues at the same average speed, how much time, to the nearest minute, will it take before the second car arrives?

Let 𝒓𝒓 represent the speed in miles per hour of the faster car, then 𝒓𝒓 − 𝟕𝟕 represents the speed in miles per hour of the slower car.

𝟓𝟓𝟏𝟏𝟐𝟐

(𝒓𝒓) + 𝟓𝟓𝟏𝟏𝟐𝟐

(𝒓𝒓 − 𝟕𝟕) = 𝟓𝟓𝟗𝟗𝟗𝟗.𝟓𝟓

𝟓𝟓𝟏𝟏𝟐𝟐

(𝒓𝒓 + 𝒓𝒓 − 𝟕𝟕) = 𝟓𝟓𝟗𝟗𝟗𝟗.𝟓𝟓

𝟓𝟓𝟏𝟏𝟐𝟐

(𝟐𝟐𝒓𝒓 − 𝟕𝟕) = 𝟓𝟓𝟗𝟗𝟗𝟗.𝟓𝟓

𝟏𝟏𝟏𝟏𝟐𝟐

(𝟐𝟐𝒓𝒓 − 𝟕𝟕) = 𝟓𝟓𝟗𝟗𝟗𝟗.𝟓𝟓

𝟐𝟐𝟏𝟏𝟏𝟏

∙𝟏𝟏𝟏𝟏𝟐𝟐

(𝟐𝟐𝒓𝒓 − 𝟕𝟕) =𝟐𝟐𝟏𝟏𝟏𝟏

∙ 𝟓𝟓𝟗𝟗𝟗𝟗.𝟓𝟓

𝟏𝟏 ∙ (𝟐𝟐𝒓𝒓 − 𝟕𝟕) =𝟏𝟏𝟏𝟏𝟗𝟗𝟗𝟗𝟏𝟏𝟏𝟏

𝟐𝟐𝒓𝒓 − 𝟕𝟕 = 𝟏𝟏𝟐𝟐𝟗𝟗 𝟐𝟐𝒓𝒓 − 𝟕𝟕+ 𝟕𝟕 = 𝟏𝟏𝟐𝟐𝟗𝟗+ 𝟕𝟕

𝟐𝟐𝒓𝒓+ 𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟔𝟔 𝟐𝟐𝒓𝒓 = 𝟏𝟏𝟏𝟏𝟔𝟔

𝟏𝟏𝟐𝟐∙ 𝟐𝟐𝒓𝒓 =

𝟏𝟏𝟐𝟐∙ 𝟏𝟏𝟏𝟏𝟔𝟔

𝟏𝟏𝒓𝒓 = 𝟓𝟓𝟏𝟏 𝒓𝒓 = 𝟓𝟓𝟏𝟏

The average speed of the faster car is 𝟓𝟓𝟏𝟏 miles per hour, so the average speed of the slower car is 𝟓𝟓𝟏𝟏 miles per hour.

𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐜𝐜𝐝𝐝 = 𝐫𝐫𝐝𝐝𝐝𝐝𝐝𝐝 ∙ 𝐝𝐝𝐝𝐝𝐜𝐜𝐝𝐝

𝒅𝒅 = 𝟓𝟓𝟏𝟏 ∙ 𝟓𝟓𝟏𝟏𝟐𝟐

𝒅𝒅 = 𝟓𝟓𝟏𝟏 ∙𝟏𝟏𝟏𝟏𝟐𝟐

𝒅𝒅 = 𝟐𝟐𝟏𝟏𝟐𝟐.𝟓𝟓

The slower car traveled 𝟐𝟐𝟏𝟏𝟐𝟐.𝟓𝟓 miles in 𝟓𝟓𝟏𝟏𝟐𝟐 hours.

𝒅𝒅 = 𝟓𝟓𝟏𝟏 ∙ 𝟓𝟓𝟏𝟏𝟐𝟐

𝒅𝒅 = 𝟓𝟓𝟏𝟏 ∙𝟏𝟏𝟏𝟏𝟐𝟐

𝒅𝒅 = 𝟑𝟑𝟏𝟏𝟗𝟗

OR

𝟓𝟓𝟗𝟗𝟗𝟗.𝟓𝟓 − 𝟐𝟐𝟏𝟏𝟐𝟐.𝟓𝟓 = 𝟑𝟑𝟏𝟏𝟗𝟗

The faster car traveled 𝟑𝟑𝟏𝟏𝟗𝟗 miles in 𝟓𝟓𝟏𝟏𝟐𝟐 hours.

The slower car traveled 𝟐𝟐𝟏𝟏𝟐𝟐.𝟓𝟓 miles in 𝟓𝟓𝟏𝟏𝟐𝟐 hours. The remainder of their trip is 𝟑𝟑𝟏𝟏.𝟓𝟓 miles because 𝟑𝟑𝟏𝟏𝟗𝟗 − 𝟐𝟐𝟏𝟏𝟐𝟐.𝟓𝟓 = 𝟑𝟑𝟏𝟏.𝟓𝟓.

𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐜𝐜𝐝𝐝 = 𝐫𝐫𝐝𝐝𝐝𝐝𝐝𝐝 ∙ 𝐝𝐝𝐝𝐝𝐜𝐜𝐝𝐝 𝟑𝟑𝟏𝟏.𝟓𝟓 = 𝟓𝟓𝟏𝟏 (𝒕𝒕)

𝟏𝟏𝟓𝟓𝟏𝟏

(𝟑𝟑𝟏𝟏.𝟓𝟓) =𝟏𝟏𝟓𝟓𝟏𝟏

(𝟓𝟓𝟏𝟏)(𝒕𝒕)

𝟑𝟑𝟏𝟏.𝟓𝟓𝟓𝟓𝟏𝟏

= 𝟏𝟏𝒕𝒕

𝟕𝟕𝟕𝟕𝟏𝟏𝟐𝟐𝟐𝟐

= 𝒕𝒕

This time is in hours. To convert to minutes, multiply by 𝟔𝟔𝟐𝟐 minutes per hour. 𝟕𝟕𝟕𝟕𝟏𝟏𝟐𝟐𝟐𝟐

∙ 𝟔𝟔𝟐𝟐 =𝟕𝟕𝟕𝟕𝟓𝟓𝟏𝟏

∙ 𝟑𝟑𝟐𝟐 =𝟐𝟐𝟑𝟑𝟏𝟏𝟐𝟐𝟓𝟓𝟏𝟏

≈ 𝟒𝟒𝟓𝟓

The slower car will arrive approximately 𝟒𝟒𝟓𝟓 minutes after the first.

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7•3 Lesson 9


11. Emily counts the triangles and parallelograms in an art piece and determines that altogether, there are 𝟒𝟒𝟐𝟐 triangles and parallelograms. If there are 𝟏𝟏𝟓𝟓𝟐𝟐 total sides, how many triangles and parallelograms are there?

If 𝒕𝒕 represents the number of triangles that Emily counted, then 𝟒𝟒𝟐𝟐 − 𝒕𝒕 represents the number of parallelograms that she counted.

𝟑𝟑𝒕𝒕 + 𝟒𝟒(𝟒𝟒𝟐𝟐 − 𝒕𝒕) = 𝟏𝟏𝟓𝟓𝟐𝟐

𝟑𝟑𝒕𝒕 + 𝟒𝟒�𝟒𝟒𝟐𝟐+ (−𝒕𝒕)� = 𝟏𝟏𝟓𝟓𝟐𝟐

𝟑𝟑𝒕𝒕 + 𝟒𝟒(𝟒𝟒𝟐𝟐) + 𝟒𝟒(−𝒕𝒕) = 𝟏𝟏𝟓𝟓𝟐𝟐

𝟑𝟑𝒕𝒕 + 𝟏𝟏𝟔𝟔𝟏𝟏+ (−𝟒𝟒𝒕𝒕) = 𝟏𝟏𝟓𝟓𝟐𝟐 𝟑𝟑𝒕𝒕 + (−𝟒𝟒𝒕𝒕) + 𝟏𝟏𝟔𝟔𝟏𝟏 = 𝟏𝟏𝟓𝟓𝟐𝟐

−𝒕𝒕+ 𝟏𝟏𝟔𝟔𝟏𝟏 = 𝟏𝟏𝟓𝟓𝟐𝟐 −𝒕𝒕+ 𝟏𝟏𝟔𝟔𝟏𝟏 − 𝟏𝟏𝟔𝟔𝟏𝟏 = 𝟏𝟏𝟓𝟓𝟐𝟐 − 𝟏𝟏𝟔𝟔𝟏𝟏

−𝒕𝒕 + 𝟐𝟐 = −𝟏𝟏𝟏𝟏 −𝒕𝒕 = −𝟏𝟏𝟏𝟏

−𝟏𝟏 ∙ (−𝒕𝒕) = −𝟏𝟏 ∙ (−𝟏𝟏𝟏𝟏) 𝟏𝟏𝒕𝒕 = 𝟏𝟏𝟏𝟏

𝒕𝒕 = 𝟏𝟏𝟏𝟏

There are 𝟏𝟏𝟏𝟏 triangles and 𝟐𝟐𝟒𝟒 parallelograms.

Note to the Teacher: Problems 12 and 13 are more difficult and may not be suitable to assign to all students to solve independently.

12. Stefan is three years younger than his sister Katie. The sum of Stefan’s age 𝟑𝟑 years ago and 𝟐𝟐𝟑𝟑 of Katie’s age at that time is 𝟏𝟏𝟐𝟐. How old is Katie now?

If 𝒔𝒔 represents Stefan’s age in years, then 𝒔𝒔 + 𝟑𝟑 represents Katie’s current age, 𝒔𝒔 − 𝟑𝟑 represents Stefan’s age 𝟑𝟑 years ago, and 𝒔𝒔 also represents Katie’s age 𝟑𝟑 years ago.

(𝒔𝒔 − 𝟑𝟑) + �𝟐𝟐𝟑𝟑� 𝒔𝒔 = 𝟏𝟏𝟐𝟐

𝒔𝒔 + (−𝟑𝟑) +𝟐𝟐𝟑𝟑𝒔𝒔 = 𝟏𝟏𝟐𝟐

𝒔𝒔 +𝟐𝟐𝟑𝟑𝒔𝒔 + (−𝟑𝟑) = 𝟏𝟏𝟐𝟐

𝟑𝟑𝟑𝟑𝒔𝒔 +

𝟐𝟐𝟑𝟑𝒔𝒔 + (−𝟑𝟑) = 𝟏𝟏𝟐𝟐

𝟓𝟓𝟑𝟑𝒔𝒔 + (−𝟑𝟑) = 𝟏𝟏𝟐𝟐

𝟓𝟓𝟑𝟑𝒔𝒔 + (−𝟑𝟑) + 𝟑𝟑 = 𝟏𝟏𝟐𝟐 + 𝟑𝟑

𝟓𝟓𝟑𝟑𝒔𝒔 + 𝟐𝟐 = 𝟏𝟏𝟓𝟓

𝟓𝟓𝟑𝟑𝒔𝒔 = 𝟏𝟏𝟓𝟓

𝟑𝟑𝟓𝟓∙𝟓𝟓𝟑𝟑𝒔𝒔 =

𝟑𝟑𝟓𝟓∙ 𝟏𝟏𝟓𝟓

𝟏𝟏𝒔𝒔 = 𝟑𝟑 ∙ 𝟑𝟑 𝒔𝒔 = 𝟗𝟗

Stefan’s current age is 𝟗𝟗 years, so Katie is currently 𝟏𝟏𝟐𝟐 years old.

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7•3 Lesson 9


13. Lucas bought a certain weight of oats for his horse at a unit price of $𝟐𝟐.𝟐𝟐𝟐𝟐 per pound. The total cost of the oats left him with $𝟏𝟏. He wanted to buy the same weight of enriched oats instead, but at $𝟐𝟐.𝟑𝟑𝟐𝟐 per pound, he would have been $𝟐𝟐 short of the total amount due. How much money did Lucas have to buy oats?

The difference in the costs is $𝟑𝟑.𝟐𝟐𝟐𝟐 for the same weight in feed.

Let 𝒘𝒘 represent the weight in pounds of feed.

𝟐𝟐.𝟑𝟑𝒘𝒘− 𝟐𝟐.𝟐𝟐𝒘𝒘 = 𝟑𝟑 𝟐𝟐.𝟏𝟏𝒘𝒘 = 𝟑𝟑 𝟏𝟏𝟏𝟏𝟐𝟐

𝒘𝒘 = 𝟑𝟑

𝟏𝟏𝟐𝟐 ∙𝟏𝟏𝟏𝟏𝟐𝟐

𝒘𝒘 = 𝟏𝟏𝟐𝟐 ∙ 𝟑𝟑

𝟏𝟏𝒘𝒘 = 𝟑𝟑𝟐𝟐 𝒘𝒘 = 𝟑𝟑𝟐𝟐

Lucas bought 𝟑𝟑𝟐𝟐 pounds of oats.

𝐂𝐂𝐂𝐂𝐝𝐝𝐝𝐝 = 𝐮𝐮𝐝𝐝𝐝𝐝𝐝𝐝 𝐩𝐩𝐫𝐫𝐝𝐝𝐜𝐜𝐝𝐝 × 𝐰𝐰𝐝𝐝𝐝𝐝𝐰𝐰𝐰𝐰𝐝𝐝 𝐂𝐂𝐂𝐂𝐝𝐝𝐝𝐝 = ($𝟐𝟐.𝟐𝟐𝟐𝟐 𝐩𝐩𝐝𝐝𝐫𝐫 𝐩𝐩𝐂𝐂𝐮𝐮𝐝𝐝𝐝𝐝) ∙ (𝟑𝟑𝟐𝟐 𝐩𝐩𝐂𝐂𝐮𝐮𝐝𝐝𝐝𝐝𝐝𝐝) 𝐂𝐂𝐂𝐂𝐝𝐝𝐝𝐝 = $𝟔𝟔.𝟐𝟐𝟐𝟐

Lucas paid $𝟔𝟔 for 𝟑𝟑𝟐𝟐 pounds of oats. Lucas had $𝟏𝟏 left after his purchase, so he started with $𝟕𝟕.

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7•3 Lesson 9


Group 1: Where can you buy a ruler that is 𝟑𝟑 feet long?

____ _____ _____ _____ _____ _____ _____ _____

3 412

3.5 −1 −2 19 18.95 4.22

What value(s) of 𝑧𝑧 makes the equation 76𝑧𝑧 +

13

= −56

true; 𝑧𝑧 = −1, 𝑧𝑧 = 2, 𝑧𝑧 = 1, or 𝑧𝑧 = − 3663? 𝐷𝐷

Find the smaller of 2 consecutive integers if the sum of the smaller and twice the larger is −4. 𝑆𝑆

Twice the sum of a number and −6 is −6. Find the number. 𝑌𝑌

Logan is 2 years older than Lindsey. Five years ago, the sum of their ages was 30. Find Lindsey’s current age. 𝐴𝐴

The total charge for a taxi ride in NYC includes an initial fee of $3.75 plus $1.25 for every 12

mile

traveled. Jodi took a taxi, and the ride cost her exactly $12.50. How many miles did she travel in the taxi?

𝑅𝑅

The perimeter of a triangular garden with 3 equal sides is 12.66 feet. What is the length of each side of the garden?

𝐸𝐸

A car travelling at 60 mph leaves Ithaca and travels west. Two hours later, a truck travelling at 55 mph leaves Elmira and travels east. Altogether, the car and truck travel 407.5 miles. How many hours does the car travel?

𝐴𝐴

The Cozo family has 5 children. While on vacation, they went to a play. They bought 5 tickets at the child’s price of $10.25 and 2 tickets at the adult’s price. If they spent a total of $89.15, how much was the price of each adult ticket?

𝐿𝐿

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7•3 Lesson 9


Group 1 Sample Solutions YARD SALE

What value(s) of 𝑧𝑧 makes the equation 76𝑧𝑧 +

13

= −56

true; 𝑧𝑧 = −1, 𝑧𝑧 = 2, 𝑧𝑧 = 1, or 𝑧𝑧 = − 3663? −𝟏𝟏

Find the smaller of 2 consecutive integers if the sum of the smaller and twice the larger is −4. −𝟐𝟐

Twice the sum of a number and −6 is −6 . Find the number. 𝟑𝟑

Logan is 2 years older than Lindsey. Five years ago, the sum of their ages was 30. Find Lindsey’s current age. 𝟏𝟏𝟗𝟗

The total charge for a taxi ride in NYC includes an initial fee of $3.75 plus $1.25 for every 12

mile

traveled. Jodi took a taxi, and the ride cost her exactly $12.50. How many miles did she travel in the taxi?

𝟑𝟑.𝟓𝟓 𝐜𝐜𝐝𝐝.

The perimeter of a triangular garden with 3 equal sides is 12.66 feet. What is the length of each side of the garden? 𝟒𝟒.𝟐𝟐𝟐𝟐 𝐟𝐟𝐝𝐝𝐝𝐝𝐝𝐝

A car travelling at 60 mph leaves Ithaca and travels west. Two hours later, a truck travelling at 55 mph leaves Elmira and travels East. Altogether, the car and truck travel 407.5 miles. How many hours does the car travel?

𝟒𝟒𝟏𝟏𝟐𝟐

𝐰𝐰𝐂𝐂𝐮𝐮𝐫𝐫𝐝𝐝

The Cozo family has 5 children. While on vacation, they went to a play. They bought 5 tickets at the child’s price of $10.25 and 2 tickets at the adult’s price. If they spent a total of $89.15, how much was the price of each adult ticket?

$𝟏𝟏𝟏𝟏.𝟗𝟗𝟓𝟓

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7•3 Lesson 9


Group 2: Where do fish keep their money?

____ _____ _____ _____ _____ _____ _____ _____ _____

2 −1 10 8 2 −6 5 50 18

What value of 𝑧𝑧 makes the equation 23𝑧𝑧 −

12

= −512

true; 𝑧𝑧 = −1, 𝑧𝑧 = 2, 𝑧𝑧 = 18, or 𝑧𝑧 = − 1

8 ? 𝐾𝐾

Find the smaller of 2 consecutive even integers if the sum of twice the smaller integer and the larger integers is −16.

𝐵𝐵

Twice the difference of a number and −3 is 4. Find the number. 𝐼𝐼

Brooke is 3 years younger than Samantha. In five years, the sum of their ages will be 29. Find Brooke’s age. 𝐸𝐸

Which of the following equations is equivalent to 4.12𝑝𝑝 + 5.2 = 8.23?

(1) 412𝑝𝑝 + 52 = 823

(2) 412𝑝𝑝 + 520 = 823

(3) 9.32𝑝𝑝 = 8.23

(4) 0.412𝑝𝑝 + 0.52 = 8.23

𝑅𝑅

The length of a rectangle is twice the width. If the perimeter of the rectangle is 30 units, find the area of the garden. 𝑁𝑁

A car traveling at 70 miles per hour traveled one hour longer than a truck traveling at 60 miles per hour. If the car and truck traveled a total of 330 miles, for how many hours did the car and truck travel altogether?

𝐴𝐴

Jeff sold half of his baseball cards then bought sixteen more. He now has 21 baseball cards. How many cards did he begin with? 𝑉𝑉

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7•3Lesson 9


Group 2 Sample Solutions RIVER BANK

What value of 𝑧𝑧 makes the equation 23 𝑧𝑧 −12 = − 5

12 true; 𝑧𝑧 = −1, 𝑧𝑧 = 2, 𝑧𝑧 = 18, or 𝑧𝑧 = − 1

8 ? 𝟏𝟏𝟏𝟏

Find the smaller of 2 consecutive even integers if the sum of twice the smaller integer and the larger integer is −16. −𝟔𝟔

Twice the difference of a number and −3 is 4. Find the number. −𝟏𝟏

Brooke is 3 years younger than Samantha. In 5 years, the sum of their ages will be 29. Find Brooke’s age. 𝟏𝟏

Which of the following equations is equivalent to 4.12𝑝𝑝 + 5.2 = 8.23?

(1) 412𝑝𝑝 + 52 = 823

(2) 412𝑝𝑝 + 520 = 823

(3) 9.32𝑝𝑝 = 8.23

(4) 0.412𝑝𝑝 + 0.52 = 8.23

𝟐𝟐

The length of a rectangle is twice the width. If the perimeter of the rectangle is 30 units, find the area of the garden. 𝟓𝟓𝟐𝟐 𝒔𝒔𝒔𝒔.𝐮𝐮𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝

A car traveling at 70 miles per hour traveled one hour longer than a truck traveling at 60 miles per hour. If the car and truck traveled a total of 330 miles, for how many hours did the car and truck travel altogether?

𝟓𝟓 𝐰𝐰𝐂𝐂𝐮𝐮𝐫𝐫𝐝𝐝

Jeff sold half of his baseball cards then bought 16 more. He now has 21 baseball cards. How many cards did he begin with? 𝟏𝟏𝟐𝟐

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7•3 Lesson 9


Group 3: The more you take, the more you leave behind. What are they?

____ _____ _____ _____ _____ _____ _____ _____ _____

8 11.93 368 156

10.50 212

356

21 4

An apple has 80 calories. This is 12 less than 14

the number of calories in a package of candy. How

many calories are in the candy? 𝑂𝑂

The ages of 3 brothers are represented by consecutive integers. If the oldest brother’s age is decreased by twice the youngest brother’s age, the result is −19. How old is the youngest brother?

𝑃𝑃

A carpenter uses 3 hinges on every door he hangs. He hangs 4 doors on the first floor and 𝑝𝑝 doors on the second floor. If he uses 36 hinges total, how many doors did he hang on the second floor? 𝐹𝐹

Kate has 12 12 pounds of chocolate. She gives each of her 5 friends 𝑝𝑝 pounds each and has 3 1

3 pounds left over. How much did she give each of her friends?

𝑇𝑇

A room is 20 feet long. If a couch that is 12 13 feet long is to be centered in the room, how big of a

table can be placed on either side of the couch? 𝐸𝐸

Which equation is equivalent to 14𝑝𝑝 +

15

= 2?

(1) 4𝑝𝑝 + 5 = 12

(2) 29𝑝𝑝 = 2

(3) 5𝑝𝑝 + 4 = 18

(4) 5𝑝𝑝 + 4 = 40

𝑆𝑆

During a recent sale, the first movie purchased cost $29, and each additional movie purchased costs 𝑚𝑚 dollars. If Jose buys 4 movies and spends a total of $64.80, how much did each additional movie cost? 𝑂𝑂

The Hipster Dance company purchases 5 bus tickets that cost $150 each, and they have 7 bags that cost 𝑏𝑏 dollars each. If the total bill is $823.50, how much does each bag cost?

𝑆𝑆

The weekend before final exams, Travis studied 1.5 hours for his science exam, 2 14 hours for his math

exam, and ℎ hours each for Spanish, English, and social studies. If he spent a total of 11 14 hours

studying, how much time did he spend studying for Spanish?

𝑇𝑇

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7•3 Lesson 9


Group 3 Sample Solutions FOOTSTEPS

An apple has 80 calories. This is 12 less than 14

the number of calories in a package of candy. How

many calories are in the candy? 𝟑𝟑𝟔𝟔𝟏𝟏 𝐜𝐜𝐝𝐝𝐜𝐜𝐂𝐂𝐫𝐫𝐝𝐝𝐝𝐝𝐝𝐝

The ages of 3 brothers are represented by consecutive integers. If the oldest brother’s age is decreased by twice the youngest brother’s age, the result is −19. How old is the youngest brother? 𝟐𝟐𝟏𝟏

A carpenter uses 3 hinges on every door he hangs. He hangs 4 doors on the first floor and 𝑝𝑝 doors on the second floor. If he uses 36 hinges total, how many doors did he hang on the second floor? 𝟏𝟏

Kate has 12 12 pounds of chocolate. She gives each of her 5 friends 𝑝𝑝 pounds each and has 3 1

3 pounds left over. How much did she give each of her friends?

𝟏𝟏𝟓𝟓𝟔𝟔

𝐩𝐩𝐂𝐂𝐮𝐮𝐝𝐝𝐝𝐝𝐝𝐝

A room is 20 feet long. If a couch that is 12 13 feet long is to be centered in the room, how big of a

table can be placed on either side of the couch? 𝟑𝟑𝟓𝟓𝟔𝟔

𝐟𝐟𝐝𝐝𝐝𝐝𝐝𝐝

Which equation is equivalent to 14𝑝𝑝 +

15

= 2?

(1) 4𝑝𝑝 + 5 = 12

(2) 29𝑝𝑝 = 2

(3) 5𝑝𝑝 + 4 = 18

(4) 5𝑝𝑝 + 4 = 40

𝟒𝟒

During a recent sale, the first movie purchased cost $29, and each additional movie purchased costs 𝑚𝑚 dollars. If Jose buys 4 movies and spends a total of $64.80, how much did each additional movie cost? $𝟏𝟏𝟏𝟏.𝟗𝟗𝟑𝟑

The Hipster Dance company purchases 5 bus tickets that cost $150 each, and they have 7 bags that cost 𝑏𝑏 dollars each. If the total bill is $823.50, how much does each bag cost? $𝟏𝟏𝟐𝟐.𝟓𝟓𝟐𝟐

The weekend before final exams, Travis studied 1.5 hours for his science exam, 2 14 hours for his math

exam, and ℎ hours each for Spanish, English, and social studies. If he spent a total of 11 14 hours

studying, how much time did he spend studying for Spanish?

𝟐𝟐𝟏𝟏𝟐𝟐

𝐰𝐰𝐂𝐂𝐮𝐮𝐫𝐫𝐝𝐝

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Lesson 10: Angle Problems and Solving Equations

7•3 Lesson 10


Student Outcomes

Students use vertical angles, adjacent angles, angles on a line, and angles at a point in a multistep problem to write and solve simple equations for an unknown angle in a figure.

Lesson Notes In Lessons 10 and 11, students apply their understanding of equations to unknown angle problems. The geometry topic is a natural context within which algebraic skills are applied. Students understand that the unknown angle is an actual, measureable angle; they simply need to find the value that makes each equation true. They set up the equations based on the angle facts they learned in Grade 4. The problems presented are not as simple as in Grade 4 because diagrams incorporate angle facts in combination, rather than in isolation. Encourage students to verify their answers by measuring relevant angles in each diagram―all diagrams are drawn to scale.

Classwork

Opening (5 minutes)

Discuss the ways in which angles are named and notated.

What do you notice about the three figures on the next page? What is the same about all three figures; what is different?

There are three angles that appear to be the same measurement but are notated differently.

What is a likely implication of the three different kinds of notation?

They indicate the different ways of labeling or identifying the angle.

Students are familiar with addressing Figure 1 as 𝑏𝑏 and having a measurement of 𝑏𝑏° and addressing Figure 2 as angle 𝐴𝐴. Elicit this from students and say that in a case like Figure 1, the angle is named by the measure of the arc, and in a case like Figure 2, the angle is named by the single letter.

In a case like Figure 3, we use three letters when we name the angle. Why use three points to name an angle?

In a figure where several angles share the same vertex, naming a particular angle by the vertex point is not sufficient information to distinguish that angle. Two additional points, one belonging to each side of the intended angle, are necessary to identify it.

Encourage students to use both multiple forms of angle notation in the table to demonstrate each angle relationship.

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Figure 1 Figure 2 Figure 3

Naming by the arc. Naming by the vertex. Naming by three points.

∠𝑏𝑏 ∠𝐴𝐴 ∠𝐶𝐶𝐴𝐴𝐶𝐶 or ∠𝐶𝐶𝐴𝐴𝐶𝐶

Recall the definitions of adjacent and vertical and the facts regarding angles on a line and angles at a point. If an abbreviation exists, students should include the abbreviation of the angle fact under the name of each relationship. In the Angle Fact column, students write the definitions and practice the different angle notations when describing the relationship in the angle fact.

Note: The angles on a line fact applies to two or more angles.

Angle Facts and Definitions

Name of Angle Relationship

Angle Fact Diagram

Adjacent Angles

Two angles, ∠𝑩𝑩𝑩𝑩𝑩𝑩 and ∠𝑩𝑩𝑩𝑩𝑪𝑪 with a common side 𝑩𝑩𝑩𝑩��⃗ , are adjacent angles if 𝑩𝑩 belongs to the interior of ∠𝑩𝑩𝑩𝑩𝑪𝑪. Angles 𝒂𝒂 and 𝒃𝒃 are adjacent angles; ∠𝑩𝑩𝑩𝑩𝑩𝑩 and ∠𝑩𝑩𝑩𝑩𝑪𝑪 are adjacent angles.

Vertical Angles (vert. ∠s)

Two angles are vertical angles (or vertically opposite angles) if their sides form two pairs of opposite rays. 𝒂𝒂 = 𝒃𝒃 𝒎𝒎∠𝑪𝑪𝑩𝑩𝑫𝑫 = 𝒎𝒎∠𝑮𝑮𝑩𝑩𝑮𝑮

Angles on a Line (∠s on a line)

The sum of the measures of two angles that share a ray and form a line is 𝟏𝟏𝟏𝟏𝟏𝟏°. 𝒂𝒂 + 𝒃𝒃 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 +𝒎𝒎∠𝑩𝑩𝑩𝑩𝑪𝑪 = 𝟏𝟏𝟏𝟏𝟏𝟏°

Angles at a Point (∠s at a point)

The measure of all angles formed by three or more rays with the same vertex is 𝟑𝟑𝟑𝟑𝟏𝟏°. 𝒂𝒂 + 𝒃𝒃 + 𝒄𝒄 = 𝟑𝟑𝟑𝟑𝟏𝟏 𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 +𝒎𝒎∠𝑩𝑩𝑩𝑩𝑪𝑪 +𝒎𝒎∠𝑪𝑪𝑩𝑩𝑩𝑩 = 𝟑𝟑𝟑𝟑𝟏𝟏°

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Use the diagram to complete the chart.

Name the angles that are …

Vertical ∠𝑩𝑩𝑮𝑮𝑩𝑩 and ∠𝑩𝑩𝑮𝑮𝑪𝑪, ∠𝑩𝑩𝑮𝑮𝑩𝑩 and ∠𝑪𝑪𝑮𝑮𝑩𝑩

Adjacent Answers include: ∠𝑩𝑩𝑮𝑮𝑩𝑩 and ∠𝑩𝑩𝑮𝑮𝑫𝑫 ∠𝑩𝑩𝑮𝑮𝑫𝑫 and ∠𝑫𝑫𝑮𝑮𝑩𝑩

Angles on a line Answers include: ∠𝑩𝑩𝑮𝑮𝑪𝑪, ∠𝑪𝑪𝑮𝑮𝑮𝑮, and ∠𝑮𝑮𝑮𝑮𝑩𝑩 ∠𝑩𝑩𝑮𝑮𝑩𝑩, ∠𝑩𝑩𝑮𝑮𝑫𝑫, and ∠𝑫𝑫𝑮𝑮𝑩𝑩

Angles at a point

∠𝑩𝑩𝑮𝑮𝑩𝑩, ∠𝑩𝑩𝑮𝑮𝑫𝑫, ∠𝑫𝑫𝑮𝑮𝑩𝑩, ∠𝑩𝑩𝑮𝑮𝑪𝑪, ∠𝑪𝑪𝑮𝑮𝑮𝑮, ∠𝑮𝑮𝑮𝑮𝑩𝑩


Students describe the angle relationship in the diagram and set up and solve an equation that models it. Have students verify their answers by measuring the unknown angle with a protractor.

Example 1

Estimate the measurement of 𝒙𝒙.

In a complete sentence, describe the angle relationship in the diagram.

∠𝑩𝑩𝑩𝑩𝑩𝑩 and ∠𝑩𝑩𝑩𝑩𝑪𝑪 are angles on a line and their measures have a sum of 𝟏𝟏𝟏𝟏𝟏𝟏°.

Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙. Then, find the measures of ∠𝑩𝑩𝑩𝑩𝑩𝑩 and confirm your answers by measuring the angle with a protractor.

𝒙𝒙 + 𝟏𝟏𝟑𝟑𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝒙𝒙 + 𝟏𝟏𝟑𝟑𝟏𝟏 − 𝟏𝟏𝟑𝟑𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟑𝟑𝟏𝟏

𝒙𝒙 = 𝟒𝟒𝟏𝟏

𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 = 𝟒𝟒𝟏𝟏°

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Exercise 1


∠𝑩𝑩𝑩𝑩𝑩𝑩, ∠𝑩𝑩𝑩𝑩𝑪𝑪, and ∠𝑪𝑪𝑩𝑩𝑮𝑮 are angles on a line and their measures have a sum of 𝟏𝟏𝟏𝟏𝟏𝟏°.

Find the measurements of ∠𝑩𝑩𝑩𝑩𝑩𝑩 and ∠𝑪𝑪𝑩𝑩𝑮𝑮.

𝟑𝟑𝒙𝒙 + 𝟗𝟗𝟏𝟏 + 𝟏𝟏𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓𝒙𝒙 + 𝟗𝟗𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏

𝟓𝟓𝒙𝒙 + 𝟗𝟗𝟏𝟏 − 𝟗𝟗𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟗𝟗𝟏𝟏

�𝟏𝟏𝟓𝟓� (𝟓𝟓𝒙𝒙) = �

𝟏𝟏𝟓𝟓� (𝟗𝟗𝟏𝟏)

𝒙𝒙 = 𝟏𝟏𝟏𝟏

𝒎𝒎∠𝑩𝑩𝑩𝑩𝑩𝑩 = 𝟑𝟑(𝟏𝟏𝟏𝟏°) = 𝟓𝟓𝟒𝟒°

𝒎𝒎∠𝑪𝑪𝑩𝑩𝑮𝑮 = 𝟏𝟏(𝟏𝟏𝟏𝟏°) = 𝟑𝟑𝟑𝟑°



Example 2


∠𝑩𝑩𝑮𝑮𝑨𝑨 and ∠𝑨𝑨𝑮𝑮𝑩𝑩 are angles on a line and their measures have a sum of 𝟏𝟏𝟏𝟏𝟏𝟏°. ∠𝑩𝑩𝑮𝑮𝑨𝑨 and ∠𝑲𝑲𝑮𝑮𝑩𝑩 are vertical angles and are of equal measurement.

Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙 and 𝒚𝒚. Find the measurements of ∠𝑨𝑨𝑮𝑮𝑩𝑩 and ∠𝑲𝑲𝑮𝑮𝑩𝑩.

𝒚𝒚 = 𝟏𝟏𝟒𝟒𝟒𝟒°; 𝒎𝒎∠𝑲𝑲𝑮𝑮𝑩𝑩 = 𝟏𝟏𝟒𝟒𝟒𝟒° (or vert. ∠s are =)

𝒙𝒙 + 𝟏𝟏𝟒𝟒𝟒𝟒 = 𝟏𝟏𝟏𝟏𝟏𝟏

𝒙𝒙 + 𝟏𝟏𝟒𝟒𝟒𝟒 − 𝟏𝟏𝟒𝟒𝟒𝟒 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟒𝟒𝟒𝟒 𝒙𝒙 = 𝟑𝟑𝟑𝟑

𝒎𝒎∠𝑨𝑨𝑮𝑮𝑩𝑩 = 𝟑𝟑𝟑𝟑°

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Exercise 2

In a complete sentence, describe the angle relationships in the diagram.

∠𝑱𝑱𝑮𝑮𝑱𝑱 and ∠𝑱𝑱𝑮𝑮𝑵𝑵 are adjacent angles and, when added together, are the measure of ∠𝑱𝑱𝑮𝑮𝑵𝑵; ∠𝑱𝑱𝑮𝑮𝑵𝑵 and ∠𝑲𝑲𝑮𝑮𝑨𝑨 are vertical angles and are of equal measurement.

Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙.

𝟑𝟑𝒙𝒙 + 𝟏𝟏𝟑𝟑 = 𝟏𝟏𝟓𝟓 𝟑𝟑𝒙𝒙 + 𝟏𝟏𝟑𝟑 − 𝟏𝟏𝟑𝟑 = 𝟏𝟏𝟓𝟓 − 𝟏𝟏𝟑𝟑

𝟑𝟑𝒙𝒙 = 𝟑𝟑𝟗𝟗

�𝟏𝟏𝟑𝟑�𝟑𝟑𝒙𝒙 = 𝟑𝟑𝟗𝟗�

𝟏𝟏𝟑𝟑�

𝒙𝒙 = 𝟏𝟏𝟑𝟑



Example 3


∠𝑮𝑮𝑲𝑲𝑮𝑮, ∠𝑮𝑮𝑲𝑲𝑫𝑫, and ∠𝑮𝑮𝑲𝑲𝑫𝑫 are angles at a point and their measures have a sum of 𝟑𝟑𝟑𝟑𝟏𝟏°.

Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙. Find the measurement of ∠𝑮𝑮𝑲𝑲𝑫𝑫 and confirm your answers by measuring the angle with a protractor.

𝒙𝒙 + 𝟗𝟗𝟏𝟏+ 𝟏𝟏𝟑𝟑𝟓𝟓 = 𝟑𝟑𝟑𝟑𝟏𝟏

𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟓𝟓 = 𝟑𝟑𝟑𝟑𝟏𝟏 𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟓𝟓 − 𝟏𝟏𝟏𝟏𝟓𝟓 = 𝟑𝟑𝟑𝟑𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟓𝟓

𝒙𝒙 = 𝟏𝟏𝟑𝟑𝟓𝟓

𝒎𝒎∠𝑮𝑮𝑲𝑲𝑫𝑫 = 𝟏𝟏𝟑𝟑𝟓𝟓°

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Exercise 3


∠𝑮𝑮𝑩𝑩𝑬𝑬, ∠𝑮𝑮𝑩𝑩𝑬𝑬, ∠𝑮𝑮𝑩𝑩𝑫𝑫, and ∠𝑫𝑫𝑩𝑩𝑮𝑮 are angles at a point and their measures sum to 𝟑𝟑𝟑𝟑𝟏𝟏°.

Find the measurement of ∠𝑮𝑮𝑩𝑩𝑬𝑬.

(𝒙𝒙 + 𝟏𝟏) + 𝟓𝟓𝟗𝟗+ 𝟏𝟏𝟏𝟏𝟑𝟑+ 𝟏𝟏𝟑𝟑𝟏𝟏 = 𝟑𝟑𝟑𝟑𝟏𝟏

𝒙𝒙 + 𝟏𝟏 + 𝟓𝟓𝟗𝟗+ 𝟏𝟏𝟏𝟏𝟑𝟑+ 𝟏𝟏𝟑𝟑𝟏𝟏 = 𝟑𝟑𝟑𝟑𝟏𝟏 𝒙𝒙 = 𝟑𝟑𝟏𝟏

𝒎𝒎∠𝑮𝑮𝑩𝑩𝑬𝑬 = (𝟑𝟑𝟏𝟏+ 𝟏𝟏)° = 𝟑𝟑𝟏𝟏°


List pairs of angles whose measurements are in a ratio of 2 ∶ 1.

Examples include: 90° and 45°, 60° and 30°, 150° and 75°. What does it mean for the ratio of the measurements of two angles to be 2 ∶ 1?

The measurement of one angle is two times the measure of the other angle. If the smaller angle is defined as 𝑥𝑥°, then the larger angle is 2𝑥𝑥°.

If the larger angle is defined as 𝑥𝑥°, then the smaller angle is 12𝑥𝑥°.

Based on the following figure, which angle relationship(s) can be utilized to find the measure of an obtuse and acute angle? Any adjacent angle pair are on a line and have a sum of 180°.

Students describe the angle relationship in the question and set up and solve an equation that models it. Have students verify their answers by measuring the unknown angle with a protractor.

Example 4

The following two lines intersect. The ratio of the measurements of the obtuse angle to the acute angle in any adjacent angle pair in this figure is 𝟏𝟏 ∶ 𝟏𝟏. In a complete sentence, describe the angle relationships in the diagram.

The measurement of an obtuse angle is twice the measurement of an acute angle in the diagram.

Scaffolding: Students may find it helpful to highlight the pairs of equal vertical angles.

MP.8

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Label the diagram with expressions that describe this relationship. Write an equation that models the angle relationship and solve for 𝒙𝒙. Find the measurements of the acute and obtuse angles.

𝟏𝟏𝒙𝒙 + 𝟏𝟏𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟏𝟏

𝟑𝟑𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟏𝟏

�𝟏𝟏𝟑𝟑� (𝟑𝟑𝒙𝒙) = �

𝟏𝟏𝟑𝟑� (𝟏𝟏𝟏𝟏𝟏𝟏)

𝒙𝒙 = 𝟑𝟑𝟏𝟏

Acute angle = 𝟑𝟑𝟏𝟏°

Obtuse angle = 𝟏𝟏𝒙𝒙° = 𝟏𝟏(𝟑𝟑𝟏𝟏°) = 𝟏𝟏𝟏𝟏𝟏𝟏°



Exercise 4

The ratio of 𝒎𝒎∠𝑮𝑮𝑫𝑫𝑬𝑬 to 𝒎𝒎∠𝑮𝑮𝑫𝑫𝑬𝑬 is 𝟏𝟏 ∶ 𝟑𝟑. In a complete sentence, describe the angle relationships in the diagram.

The measurement of ∠𝑮𝑮𝑫𝑫𝑬𝑬 is 𝟏𝟏𝟑𝟑

the measurement of ∠𝑮𝑮𝑫𝑫𝑬𝑬; The measurements of ∠𝑮𝑮𝑫𝑫𝑬𝑬 and ∠𝑮𝑮𝑫𝑫𝑬𝑬 have a sum of 𝟗𝟗𝟏𝟏°.

Find the measures of ∠𝑮𝑮𝑫𝑫𝑬𝑬 and ∠𝑮𝑮𝑫𝑫𝑬𝑬.

𝟏𝟏𝒙𝒙 + 𝟑𝟑𝒙𝒙 = 𝟗𝟗𝟏𝟏

𝟓𝟓𝒙𝒙 = 𝟗𝟗𝟏𝟏

�𝟏𝟏𝟓𝟓� (𝟓𝟓𝒙𝒙) = �

𝟏𝟏𝟓𝟓� (𝟗𝟗𝟏𝟏)

𝒙𝒙 = 𝟏𝟏𝟏𝟏

𝒎𝒎∠𝑮𝑮𝑫𝑫𝑬𝑬 = 𝟏𝟏(𝟏𝟏𝟏𝟏°) = 𝟑𝟑𝟑𝟑°

𝒎𝒎∠𝑮𝑮𝑫𝑫𝑬𝑬 = 𝟑𝟑(𝟏𝟏𝟏𝟏°) = 𝟓𝟓𝟒𝟒°

Relevant Vocabulary

ADJACENT ANGLES: Two angles ∠𝑩𝑩𝑩𝑩𝑩𝑩 and ∠𝑩𝑩𝑩𝑩𝑪𝑪 with a common side 𝑩𝑩𝑩𝑩��⃗ are adjacent angles if 𝑩𝑩 belongs to the interior of ∠𝑩𝑩𝑩𝑩𝑪𝑪.

VERTICAL ANGLES: Two angles are vertical angles (or vertically opposite angles) if their sides form two pairs of opposite rays.

ANGLES ON A LINE: The sum of the measures of adjacent angles on a line is 𝟏𝟏𝟏𝟏𝟏𝟏°.

ANGLES AT A POINT: The sum of the measures of adjacent angles at a point is 𝟑𝟑𝟑𝟑𝟏𝟏°.


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7•3 Lesson 10

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Exit Ticket In a complete sentence, describe the relevant angle relationships in the following diagram. That is, describe the angle relationships you could use to determine the value of 𝑥𝑥.

Use the angle relationships described above to write an equation to solve for 𝑥𝑥. Then, determine the measurements of ∠𝐽𝐽𝐴𝐴𝐽𝐽 and ∠𝐽𝐽𝐴𝐴𝐻𝐻.

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7•3 Lesson 10


In a complete sentence, describe the relevant angle relationships in the following diagram. That is, describe the angle relationships you could use to determine the value of 𝒙𝒙.

∠𝑲𝑲𝑩𝑩𝑮𝑮 and ∠𝑮𝑮𝑩𝑩𝑫𝑫 are adjacent angles whose measurements are equal to ∠𝑲𝑲𝑩𝑩𝑫𝑫; ∠𝑲𝑲𝑩𝑩𝑫𝑫 and ∠𝑱𝑱𝑩𝑩𝑮𝑮 are vertical angles and are of equal measurement.

Use the angle relationships described above to write an equation to solve for 𝒙𝒙. Then, determine the measurements of ∠𝑱𝑱𝑩𝑩𝑬𝑬 and ∠𝑬𝑬𝑩𝑩𝑮𝑮.

𝟓𝟓𝒙𝒙 + 𝟑𝟑𝒙𝒙 = 𝟗𝟗𝟏𝟏 + 𝟑𝟑𝟏𝟏

𝟏𝟏𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟏𝟏

�𝟏𝟏𝟏𝟏� (𝟏𝟏𝒙𝒙) = �

𝟏𝟏𝟏𝟏� (𝟏𝟏𝟏𝟏𝟏𝟏)

𝒙𝒙 = 𝟏𝟏𝟓𝟓

𝒎𝒎∠𝑱𝑱𝑩𝑩𝑬𝑬 = 𝟑𝟑(𝟏𝟏𝟓𝟓°) = 𝟒𝟒𝟓𝟓°

𝒎𝒎∠𝑬𝑬𝑩𝑩𝑮𝑮 = 𝟓𝟓(𝟏𝟏𝟓𝟓°) = 𝟏𝟏𝟓𝟓°


For each question, use angle relationships to write an equation in order to solve for each variable. Determine the indicated angles. You can check your answers by measuring each angle with a protractor.

1. In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measurement of ∠𝑪𝑪𝑩𝑩𝑮𝑮.

One possible response: ∠𝑩𝑩𝑩𝑩𝑪𝑪, ∠𝑪𝑪𝑩𝑩𝑮𝑮, and ∠𝑫𝑫𝑩𝑩𝑮𝑮 are angles on a line and their measures sum to 𝟏𝟏𝟏𝟏𝟏𝟏°.

𝟗𝟗𝟏𝟏+ 𝒙𝒙 + 𝟑𝟑𝟓𝟓 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝒙𝒙 + 𝟏𝟏𝟓𝟓𝟓𝟓 = 𝟏𝟏𝟏𝟏𝟏𝟏

𝒙𝒙 + 𝟏𝟏𝟓𝟓𝟓𝟓 − 𝟏𝟏𝟓𝟓𝟓𝟓 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟓𝟓𝟓𝟓 𝒙𝒙 = 𝟏𝟏𝟓𝟓

𝒎𝒎∠𝑪𝑪𝑩𝑩𝑮𝑮 = 𝟏𝟏𝟓𝟓°

2. In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measurement of ∠𝑸𝑸𝑸𝑸𝑸𝑸.

∠𝑸𝑸𝑸𝑸𝑸𝑸, ∠𝑸𝑸𝑸𝑸𝑹𝑹, and ∠𝑹𝑹𝑸𝑸𝑺𝑺 are angles on a line and their measures sum to 𝟏𝟏𝟏𝟏𝟏𝟏°.

𝒇𝒇 + 𝟏𝟏𝟓𝟓𝟒𝟒+ 𝒇𝒇 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝒇𝒇 + 𝟏𝟏𝟓𝟓𝟒𝟒 = 𝟏𝟏𝟏𝟏𝟏𝟏

𝟏𝟏𝒇𝒇 + 𝟏𝟏𝟓𝟓𝟒𝟒 − 𝟏𝟏𝟓𝟓𝟒𝟒 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟓𝟓𝟒𝟒 𝟏𝟏𝒇𝒇 = 𝟏𝟏𝟑𝟑

�𝟏𝟏𝟏𝟏�𝟏𝟏𝒇𝒇 = �

𝟏𝟏𝟏𝟏�𝟏𝟏𝟑𝟑

𝒇𝒇 = 𝟏𝟏𝟑𝟑

𝒎𝒎∠𝑸𝑸𝑸𝑸𝑸𝑸 = 𝟏𝟏𝟑𝟑°

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3. In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measurements of ∠𝑩𝑩𝑸𝑸𝑪𝑪 and ∠𝑮𝑮𝑸𝑸𝑫𝑫.

∠𝑩𝑩𝑸𝑸𝑩𝑩, ∠𝑩𝑩𝑸𝑸𝑪𝑪, ∠𝑪𝑪𝑸𝑸𝑮𝑮, ∠𝑮𝑮𝑸𝑸𝑫𝑫, and ∠𝑫𝑫𝑸𝑸𝑮𝑮 are angles on a line and their measures sum to 𝟏𝟏𝟏𝟏𝟏𝟏°.

𝟏𝟏𝟏𝟏 + 𝟏𝟏𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟑𝟑 + 𝟑𝟑𝒙𝒙 + 𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟓𝟓 = 𝟏𝟏𝟏𝟏𝟏𝟏

𝟓𝟓𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟓𝟓 − 𝟏𝟏𝟏𝟏𝟓𝟓 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟓𝟓 𝟓𝟓𝒙𝒙 = 𝟓𝟓𝟓𝟓

�𝟏𝟏𝟓𝟓�𝟓𝟓𝒙𝒙 = �

𝟏𝟏𝟓𝟓�𝟓𝟓𝟓𝟓

𝒙𝒙 = 𝟏𝟏𝟏𝟏

𝒎𝒎∠𝑩𝑩𝑸𝑸𝑪𝑪 = 𝟏𝟏(𝟏𝟏𝟏𝟏°) = 𝟏𝟏𝟏𝟏°

𝒎𝒎∠𝑮𝑮𝑸𝑸𝑫𝑫 = 𝟑𝟑(𝟏𝟏𝟏𝟏°) = 𝟑𝟑𝟑𝟑°

4. In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measure of 𝒙𝒙.

All of the angles in the diagram are angles at a point and their measures sum to 𝟑𝟑𝟑𝟑𝟏𝟏°.

𝟒𝟒(𝒙𝒙 + 𝟏𝟏𝟏𝟏) = 𝟑𝟑𝟑𝟑𝟏𝟏 𝟒𝟒𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟒𝟒 = 𝟑𝟑𝟑𝟑𝟏𝟏

𝟒𝟒𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟒𝟒 − 𝟏𝟏𝟏𝟏𝟒𝟒 = 𝟑𝟑𝟑𝟑𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟒𝟒 𝟒𝟒𝒙𝒙 = 𝟏𝟏𝟑𝟑

�𝟏𝟏𝟒𝟒�𝟒𝟒𝒙𝒙 = �

𝟏𝟏𝟒𝟒�𝟏𝟏𝟑𝟑

𝒙𝒙 = 𝟏𝟏𝟗𝟗

5. In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measures of 𝒙𝒙 and 𝒚𝒚.

∠𝑩𝑩𝑲𝑲𝑮𝑮, ∠𝑮𝑮𝑲𝑲𝑪𝑪, and ∠𝑪𝑪𝑲𝑲𝑩𝑩 are angles on a line and their measures sum to 𝟏𝟏𝟏𝟏𝟏𝟏°. Since ∠𝑫𝑫𝑲𝑲𝑩𝑩 and ∠𝑩𝑩𝑲𝑲𝑮𝑮 form a straight angle and the measurement of ∠𝑫𝑫𝑲𝑲𝑩𝑩 is 𝟗𝟗𝟏𝟏°, ∠𝑩𝑩𝑲𝑲𝑮𝑮 is 𝟗𝟗𝟏𝟏°, making ∠𝑩𝑩𝑲𝑲𝑮𝑮 and ∠𝑩𝑩𝑲𝑲𝑩𝑩 form a right angle and their measures have a sum of 𝟗𝟗𝟏𝟏°.

𝒙𝒙 + 𝟏𝟏𝟓𝟓+ 𝟗𝟗𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟓𝟓 = 𝟏𝟏𝟏𝟏𝟏𝟏

𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟓𝟓 − 𝟏𝟏𝟏𝟏𝟓𝟓 = 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟓𝟓 𝒙𝒙 = 𝟑𝟑𝟓𝟓

(𝟑𝟑𝟓𝟓) + 𝒚𝒚 = 𝟗𝟗𝟏𝟏 𝟑𝟑𝟓𝟓 − 𝟑𝟑𝟓𝟓 + 𝒚𝒚 = 𝟗𝟗𝟏𝟏 − 𝟑𝟑𝟓𝟓

𝒚𝒚 = 𝟏𝟏𝟓𝟓

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6. In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measures of 𝒙𝒙 and 𝒚𝒚.

∠𝑮𝑮𝑩𝑩𝑮𝑮 and ∠𝑫𝑫𝑩𝑩𝑲𝑲 are vertical angles and are of equal measurement. ∠𝑮𝑮𝑩𝑩𝑮𝑮 and ∠𝑮𝑮𝑩𝑩𝑪𝑪 form a right angle and their measures have a sum of 𝟗𝟗𝟏𝟏°.

𝟏𝟏𝒙𝒙 + 𝟏𝟏𝟒𝟒 = 𝟗𝟗𝟏𝟏 𝟏𝟏𝒙𝒙 + 𝟏𝟏𝟒𝟒 − 𝟏𝟏𝟒𝟒 = 𝟗𝟗𝟏𝟏 − 𝟏𝟏𝟒𝟒

𝟏𝟏𝒙𝒙 = 𝟑𝟑𝟑𝟑

�𝟏𝟏𝟏𝟏�𝟏𝟏𝒙𝒙 = �

𝟏𝟏𝟏𝟏�𝟑𝟑𝟑𝟑

𝒙𝒙 = 𝟑𝟑𝟑𝟑

𝟑𝟑𝒚𝒚 = 𝟑𝟑𝟑𝟑

�𝟏𝟏𝟑𝟑�𝟑𝟑𝒚𝒚 = �

𝟏𝟏𝟑𝟑�𝟑𝟑𝟑𝟑

𝒚𝒚 = 𝟏𝟏𝟏𝟏

7. In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measures of ∠𝑩𝑩𝑩𝑩𝑪𝑪 and ∠𝑪𝑪𝑩𝑩𝑮𝑮.

∠𝑩𝑩𝑩𝑩𝑪𝑪 and ∠𝑪𝑪𝑩𝑩𝑮𝑮 form a right angle and their measures have a sum of 𝟗𝟗𝟏𝟏°.

�𝟑𝟑𝟏𝟏𝒙𝒙 + 𝟏𝟏𝟏𝟏�+ 𝟏𝟏𝒙𝒙 = 𝟗𝟗𝟏𝟏

𝟏𝟏𝟏𝟏𝒙𝒙 + 𝟏𝟏𝟏𝟏 = 𝟗𝟗𝟏𝟏

𝟏𝟏𝟏𝟏𝒙𝒙 + 𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏 = 𝟗𝟗𝟏𝟏 − 𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏𝒙𝒙 = 𝟏𝟏𝟏𝟏

�𝟏𝟏𝟏𝟏�𝟏𝟏𝟏𝟏𝒙𝒙 = 𝟏𝟏𝟏𝟏�

𝟏𝟏𝟏𝟏�

𝒙𝒙 = 𝟏𝟏𝟏𝟏

𝒎𝒎∠𝑩𝑩𝑩𝑩𝑪𝑪 =𝟑𝟑𝟏𝟏

(𝟏𝟏𝟏𝟏°) + 𝟏𝟏𝟏𝟏° = 𝟓𝟓𝟏𝟏°

𝒎𝒎∠𝑪𝑪𝑩𝑩𝑮𝑮 = 𝟏𝟏(𝟏𝟏𝟏𝟏°) = 𝟒𝟒𝟏𝟏°

8. In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measure of ∠𝑩𝑩𝑸𝑸𝑮𝑮.

∠𝑪𝑪𝑸𝑸𝑮𝑮 and ∠𝑩𝑩𝑸𝑸𝑫𝑫 are vertical angles and are of equal measurement. ∠𝑩𝑩𝑸𝑸𝑮𝑮 and ∠𝑮𝑮𝑸𝑸𝑫𝑫 are adjacent angles and their measures sum to the measure of ∠𝑩𝑩𝑸𝑸𝑫𝑫.

𝟑𝟑𝒙𝒙 + 𝟓𝟓𝟑𝟑 = 𝟏𝟏𝟓𝟓𝟓𝟓

𝟑𝟑𝒙𝒙 + 𝟓𝟓𝟑𝟑 − 𝟓𝟓𝟑𝟑 = 𝟏𝟏𝟓𝟓𝟓𝟓 − 𝟓𝟓𝟑𝟑 𝟑𝟑𝒙𝒙 = 𝟗𝟗𝟗𝟗

�𝟏𝟏𝟑𝟑�𝟑𝟑𝒙𝒙 = �

𝟏𝟏𝟑𝟑�𝟗𝟗𝟗𝟗

𝒙𝒙 = 𝟑𝟑𝟑𝟑

𝒎𝒎∠𝑩𝑩𝑸𝑸𝑮𝑮 = 𝟑𝟑(𝟑𝟑𝟑𝟑°) = 𝟗𝟗𝟗𝟗°

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9. The ratio of the measures of a pair of adjacent angles on a line is 𝟒𝟒 ∶ 𝟓𝟓.

a. Find the measures of the two angles.

∠𝟏𝟏 = 𝟒𝟒𝒙𝒙, ∠𝟏𝟏 = 𝟓𝟓𝒙𝒙

𝟒𝟒𝒙𝒙 + 𝟓𝟓𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝟗𝟗𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟏𝟏

�𝟏𝟏𝟗𝟗�𝟗𝟗𝒙𝒙 = �

𝟏𝟏𝟗𝟗�𝟏𝟏𝟏𝟏𝟏𝟏

𝒙𝒙 = 𝟏𝟏𝟏𝟏

∠𝟏𝟏 = 𝟒𝟒(𝟏𝟏𝟏𝟏°) = 𝟏𝟏𝟏𝟏°

∠𝟏𝟏 = 𝟓𝟓(𝟏𝟏𝟏𝟏°) = 𝟏𝟏𝟏𝟏𝟏𝟏°

b. Draw a diagram to scale of these adjacent angles. Indicate the measurements of each angle.

10. The ratio of the measures of three adjacent angles on a line is 𝟑𝟑 ∶ 𝟒𝟒 ∶ 𝟓𝟓.

a. Find the measures of the three angles.

∠𝟏𝟏 = 𝟑𝟑𝒙𝒙, ∠𝟏𝟏 = 𝟒𝟒𝒙𝒙, ∠𝟑𝟑 = 𝟓𝟓𝒙𝒙

𝟑𝟑𝒙𝒙 + 𝟒𝟒𝒙𝒙 + 𝟓𝟓𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟏𝟏

�𝟏𝟏𝟏𝟏𝟏𝟏�𝟏𝟏𝟏𝟏𝒙𝒙 = �

𝟏𝟏𝟏𝟏𝟏𝟏�𝟏𝟏𝟏𝟏𝟏𝟏

𝒙𝒙 = 𝟏𝟏𝟓𝟓

∠𝟏𝟏 = 𝟑𝟑(𝟏𝟏𝟓𝟓°) = 𝟒𝟒𝟓𝟓°

∠𝟏𝟏 = 𝟒𝟒(𝟏𝟏𝟓𝟓°) = 𝟑𝟑𝟏𝟏°

∠𝟑𝟑 = 𝟓𝟓(𝟏𝟏𝟓𝟓°) = 𝟏𝟏𝟓𝟓°

b. Draw a diagram to scale of these adjacent angles. Indicate the measurements of each angle.

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7•3 Lesson 11



Student Outcomes

Students use vertical angles, adjacent angles, angles on a line, and angles at a point in a multi-step problem to write and solve simple equations for an unknown angle in a figure.

Lesson Notes Lesson 11 continues where Lesson 10 ended and incorporates slightly more difficult problems. At the heart of each problem is the need to model the angle relationships in an equation and then solve for the unknown angle. The diagrams are all drawn to scale; students should verify their answers by using a protractor to measure relevant angles.

Classwork



Opening Exercise

a. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙. Confirm your answers by measuring the angle with a protractor.

The angles marked by 𝒙𝒙°, 𝟗𝟗𝟗𝟗°, and 𝟏𝟏𝟏𝟏° are angles on a line and have a sum of 𝟏𝟏𝟏𝟏𝟗𝟗°.

𝒙𝒙 + 𝟗𝟗𝟗𝟗 + 𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟗𝟗 𝒙𝒙 + 𝟏𝟏𝟗𝟗𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟗𝟗

𝒙𝒙 + 𝟏𝟏𝟗𝟗𝟏𝟏 − 𝟏𝟏𝟗𝟗𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟗𝟗 − 𝟏𝟏𝟗𝟗𝟏𝟏 𝒙𝒙 = 𝟕𝟕𝟕𝟕

b. 𝑪𝑪𝑪𝑪�⃖��⃗ and 𝑬𝑬𝑬𝑬�⃖��⃗ are intersecting lines. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for 𝒚𝒚. Confirm your answers by measuring the angle with a protractor.

The adjacent angles marked by 𝒚𝒚° and 𝟓𝟓𝟏𝟏° together form the angle that is vertically opposite and equal to the angle measuring 𝟏𝟏𝟏𝟏𝟕𝟕°.

𝒚𝒚 + 𝟓𝟓𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟕𝟕

𝒚𝒚+ 𝟓𝟓𝟏𝟏 − 𝟓𝟓𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟕𝟕 − 𝟓𝟓𝟏𝟏 𝒚𝒚 = 𝟗𝟗𝟕𝟕

Note to Teacher: You may choose or offer a choice of difficulty to students in the Opening Exercise.

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c. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for 𝒃𝒃. Confirm your answers by measuring the angle with a protractor.

The adjacent angles marked by 𝟓𝟓𝟗𝟗°, 𝟏𝟏𝟏𝟏°, 𝒃𝒃°, 𝟕𝟕𝟓𝟓°, and 𝟗𝟗𝟗𝟗° are angles at a point and together have a sum of 𝟑𝟑𝟕𝟕𝟗𝟗°.

𝟓𝟓𝟗𝟗+ 𝟏𝟏𝟏𝟏 + 𝒃𝒃 + 𝟕𝟕𝟓𝟓 + 𝟗𝟗𝟗𝟗 = 𝟑𝟑𝟕𝟕𝟗𝟗

𝒃𝒃 + 𝟐𝟐𝟓𝟓𝟓𝟓 = 𝟑𝟑𝟕𝟕𝟗𝟗 − 𝟐𝟐𝟓𝟓𝟓𝟓 𝒃𝒃 = 𝟏𝟏𝟗𝟗𝟓𝟓

d. The following figure shows three lines intersecting at a point. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for 𝒛𝒛. Confirm your answers by measuring the angle with a protractor.

The angles marked by 𝒛𝒛°, 𝟏𝟏𝟓𝟓𝟏𝟏°, and 𝒛𝒛° are angles on a line and have a sum of 𝟏𝟏𝟏𝟏𝟗𝟗°.

𝒛𝒛 + 𝟏𝟏𝟓𝟓𝟏𝟏+ 𝒛𝒛 = 𝟏𝟏𝟏𝟏𝟗𝟗 𝟐𝟐𝒛𝒛 + 𝟏𝟏𝟓𝟓𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟗𝟗

𝟐𝟐𝒛𝒛 + 𝟏𝟏𝟓𝟓𝟏𝟏 − 𝟏𝟏𝟓𝟓𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟗𝟗 − 𝟏𝟏𝟓𝟓𝟏𝟏 𝟐𝟐𝒛𝒛 = 𝟐𝟐𝟐𝟐

𝒛𝒛 = 𝟏𝟏𝟏𝟏

e. Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙. In a complete sentence, describe the angle relationship in the diagram. Find the measurements of ∠𝑬𝑬𝑬𝑬𝑬𝑬 and ∠𝑪𝑪𝑬𝑬𝑪𝑪. Confirm your answers by measuring the angle with a protractor.

∠𝑪𝑪𝑬𝑬𝑪𝑪, ∠𝑪𝑪𝑬𝑬𝑬𝑬, and ∠𝑬𝑬𝑬𝑬𝑬𝑬 are angles on a line and their measures have a sum of 𝟏𝟏𝟏𝟏𝟗𝟗°.

𝟓𝟓𝒙𝒙 + 𝟗𝟗𝟗𝟗+ 𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟗𝟗 𝟕𝟕𝒙𝒙 + 𝟗𝟗𝟗𝟗 = 𝟏𝟏𝟏𝟏𝟗𝟗

𝟕𝟕𝒙𝒙 + 𝟗𝟗𝟗𝟗 − 𝟗𝟗𝟗𝟗 = 𝟏𝟏𝟏𝟏𝟗𝟗 − 𝟗𝟗𝟗𝟗 𝟕𝟕𝒙𝒙 = 𝟗𝟗𝟗𝟗

�𝟏𝟏𝟕𝟕�𝟕𝟕𝒙𝒙 = �

𝟏𝟏𝟕𝟕�𝟗𝟗𝟗𝟗

𝒙𝒙 = 𝟏𝟏𝟓𝟓

∠𝑬𝑬𝑬𝑬𝑬𝑬 = 𝟏𝟏𝟓𝟓°

∠𝑪𝑪𝑬𝑬𝑪𝑪 = 𝟓𝟓(𝟏𝟏𝟓𝟓°) = 𝟕𝟕𝟓𝟓°

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Example 1

The following figure shows three lines intersecting at a point. In a complete sentence, describe the angle relationship in the diagram. Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙. Confirm your answers by measuring the angle with a protractor. The angles 𝟏𝟏𝟕𝟕°, 𝟕𝟕𝟏𝟏°, and the angle between them, which is vertically opposite and equal in measure to 𝒙𝒙, are angles on a line and have a sum of 𝟏𝟏𝟏𝟏𝟗𝟗°.

𝟏𝟏𝟕𝟕+ 𝒙𝒙 + 𝟕𝟕𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟗𝟗

𝒙𝒙 + 𝟏𝟏𝟓𝟓𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟗𝟗 𝒙𝒙 + 𝟏𝟏𝟓𝟓𝟏𝟏 − 𝟏𝟏𝟓𝟓𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟗𝟗 − 𝟏𝟏𝟓𝟓𝟏𝟏

𝒙𝒙 = 𝟐𝟐𝟕𝟕


The following figure shows four lines intersecting at a point. In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙 and 𝒚𝒚. Confirm your answers by measuring the angle with a protractor.

The angles 𝒙𝒙°, 𝟐𝟐𝟓𝟓°, 𝒚𝒚°, and 𝟏𝟏𝟗𝟗° are angles on a line and have a sum of 𝟏𝟏𝟏𝟏𝟗𝟗°; the angle marked 𝒚𝒚° is vertically opposite and equal to 𝟗𝟗𝟕𝟕°.

𝒚𝒚 = 𝟗𝟗𝟕𝟕, vert. ∠s

𝒙𝒙 + 𝟐𝟐𝟓𝟓+ (𝟗𝟗𝟕𝟕) + 𝟏𝟏𝟗𝟗 = 𝟏𝟏𝟏𝟏𝟗𝟗 𝒙𝒙 + 𝟏𝟏𝟕𝟕𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟗𝟗

𝒙𝒙 + 𝟏𝟏𝟕𝟕𝟏𝟏 − 𝟏𝟏𝟕𝟕𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟗𝟗 − 𝟏𝟏𝟕𝟕𝟏𝟏 𝒙𝒙 = 𝟏𝟏𝟗𝟗


Example 2

In a complete sentence, describe the angle relationships in the diagram. You may label the diagram to help describe the angle relationships. Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙. Confirm your answers by measuring the angle with a protractor.

The angle formed by adjacent angles 𝒂𝒂° and 𝒃𝒃° is vertically opposite to the 𝟕𝟕𝟕𝟕° angle. The angles 𝒙𝒙°, 𝒂𝒂°, and 𝒃𝒃° are adjacent angles that have a sum of 𝟗𝟗𝟗𝟗° (since the adjacent angle is a right angle and together the angles are on a line).

𝒙𝒙 + 𝟕𝟕𝟕𝟕 = 𝟗𝟗𝟗𝟗 𝒙𝒙 + 𝟕𝟕𝟕𝟕 − 𝟕𝟕𝟕𝟕 = 𝟗𝟗𝟗𝟗 − 𝟕𝟕𝟕𝟕

𝒙𝒙 = 𝟏𝟏𝟑𝟑

𝒂𝒂˚

𝒃𝒃˚

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In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙 and 𝒚𝒚. Confirm your answers by measuring the angles with a protractor.

The measures of angles 𝒙𝒙 and 𝒚𝒚 have a sum of 𝟗𝟗𝟗𝟗°; the measures of angles 𝒙𝒙 and 𝟐𝟐𝟕𝟕 have a sum of 𝟗𝟗𝟗𝟗°.

𝒙𝒙 + 𝟐𝟐𝟕𝟕 = 𝟗𝟗𝟗𝟗 𝒙𝒙 + 𝟐𝟐𝟕𝟕 − 𝟐𝟐𝟕𝟕 = 𝟗𝟗𝟗𝟗 − 𝟐𝟐𝟕𝟕

𝒙𝒙 = 𝟕𝟕𝟑𝟑

(𝟕𝟕𝟑𝟑) + 𝒚𝒚 = 𝟗𝟗𝟗𝟗 𝟕𝟕𝟑𝟑 − 𝟕𝟕𝟑𝟑 + 𝒚𝒚 = 𝟗𝟗𝟗𝟗 − 𝟕𝟕𝟑𝟑

𝒚𝒚 = 𝟐𝟐𝟕𝟕


Example 3

In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙. Find the measures of ∠𝑱𝑱𝑪𝑪𝑱𝑱 and ∠𝑮𝑮𝑪𝑪𝑬𝑬. Confirm your answers by measuring the angle with a protractor.

The sum of the degree measurements of ∠𝑱𝑱𝑪𝑪𝑱𝑱, ∠𝑮𝑮𝑪𝑪𝑱𝑱, ∠𝑮𝑮𝑪𝑪𝑬𝑬, and the arc that subtends ∠𝑱𝑱𝑪𝑪𝑬𝑬 is 𝟑𝟑𝟕𝟕𝟗𝟗°.

𝟐𝟐𝟐𝟐𝟓𝟓 + 𝟐𝟐𝒙𝒙 + 𝟗𝟗𝟗𝟗 + 𝟑𝟑𝒙𝒙 = 𝟑𝟑𝟕𝟕𝟗𝟗 𝟑𝟑𝟏𝟏𝟓𝟓 + 𝟓𝟓𝒙𝒙 = 𝟑𝟑𝟕𝟕𝟗𝟗

𝟑𝟑𝟏𝟏𝟓𝟓 − 𝟑𝟑𝟏𝟏𝟓𝟓 + 𝟓𝟓𝒙𝒙 = 𝟑𝟑𝟕𝟕𝟗𝟗 − 𝟑𝟑𝟏𝟏𝟓𝟓 𝟓𝟓𝒙𝒙 = 𝟏𝟏𝟓𝟓

�𝟏𝟏𝟓𝟓�𝟓𝟓𝒙𝒙 = �

𝟏𝟏𝟓𝟓�𝟏𝟏𝟓𝟓

𝒙𝒙 = 𝟗𝟗

𝒎𝒎∠𝑱𝑱𝑪𝑪𝑱𝑱 = 𝟐𝟐(𝟗𝟗°) = 𝟏𝟏𝟏𝟏° 𝒎𝒎∠𝑮𝑮𝑪𝑪𝑬𝑬 = 𝟑𝟑(𝟗𝟗°) = 𝟐𝟐𝟕𝟕°


In a complete sentence, describe the angle relationships in the diagram. Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙. Find the measure of ∠𝑱𝑱𝑱𝑱𝑮𝑮. Confirm your answer by measuring the angle with a protractor.

The sum of the degree measurements of ∠𝑳𝑳𝑱𝑱𝑱𝑱, ∠𝑱𝑱𝑱𝑱𝑮𝑮, ∠𝑮𝑮𝑱𝑱𝑮𝑮, and the arc that subtends ∠𝑳𝑳𝑱𝑱𝑮𝑮 is 𝟑𝟑𝟕𝟕𝟗𝟗°.

𝟓𝟓𝒙𝒙 + 𝟐𝟐𝟏𝟏 + 𝒙𝒙 + 𝟗𝟗𝟗𝟗 = 𝟑𝟑𝟕𝟕𝟗𝟗 𝟕𝟕𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟑𝟑𝟕𝟕𝟗𝟗

𝟕𝟕𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟑𝟑𝟕𝟕𝟗𝟗 − 𝟏𝟏𝟏𝟏𝟏𝟏 𝟕𝟕𝒙𝒙 = 𝟐𝟐𝟏𝟏𝟕𝟕

�𝟏𝟏𝟕𝟕�𝟕𝟕𝒙𝒙 = �

𝟏𝟏𝟕𝟕�𝟐𝟐𝟏𝟏𝟕𝟕

𝒙𝒙 = 𝟏𝟏𝟏𝟏 𝒎𝒎∠𝑱𝑱𝑱𝑱𝑮𝑮 = 𝟏𝟏𝟏𝟏°

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Example 4

In the accompanying diagram, the measure of ∠𝑪𝑪𝑬𝑬𝑬𝑬 is four times the measure of ∠𝑬𝑬𝑬𝑬𝑮𝑮.

a. Label ∠𝑪𝑪𝑬𝑬𝑬𝑬 as 𝒚𝒚° and ∠𝑬𝑬𝑬𝑬𝑮𝑮 as 𝒙𝒙°. Write an equation that describes the relationship between ∠𝑪𝑪𝑬𝑬𝑬𝑬 and ∠𝑬𝑬𝑬𝑬𝑮𝑮.

𝒚𝒚 = 𝟏𝟏𝒙𝒙

b. Find the value of 𝒙𝒙.

𝟓𝟓𝟗𝟗 + 𝒙𝒙 + 𝟏𝟏𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟗𝟗

𝟓𝟓𝟗𝟗 + 𝟓𝟓𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟗𝟗 𝟓𝟓𝒙𝒙 + 𝟓𝟓𝟗𝟗 − 𝟓𝟓𝟗𝟗 = 𝟏𝟏𝟏𝟏𝟗𝟗 − 𝟓𝟓𝟗𝟗

𝟓𝟓𝒙𝒙 = 𝟏𝟏𝟑𝟑𝟗𝟗

�𝟏𝟏𝟓𝟓� (𝟓𝟓𝒙𝒙) = �

𝟏𝟏𝟓𝟓� (𝟏𝟏𝟑𝟑𝟗𝟗)

𝒙𝒙 = 𝟐𝟐𝟕𝟕

c. Find the measures of ∠𝑬𝑬𝑬𝑬𝑮𝑮, ∠𝑪𝑪𝑬𝑬𝑪𝑪, ∠𝑪𝑪𝑬𝑬𝑬𝑬, ∠𝑮𝑮𝑬𝑬𝑬𝑬, and ∠𝑪𝑪𝑬𝑬𝑬𝑬.

𝒎𝒎∠𝑬𝑬𝑬𝑬𝑮𝑮 = 𝟐𝟐𝟕𝟕°

𝒎𝒎∠𝑪𝑪𝑬𝑬𝑪𝑪 = 𝟐𝟐𝟕𝟕°

𝒎𝒎∠𝑪𝑪𝑬𝑬𝑬𝑬 = 𝟏𝟏(𝟐𝟐𝟕𝟕°) = 𝟏𝟏𝟗𝟗𝟏𝟏°

𝒎𝒎∠𝑮𝑮𝑬𝑬𝑬𝑬 = 𝟓𝟓𝟗𝟗°

𝒎𝒎∠𝑪𝑪𝑬𝑬𝑬𝑬 = 𝟏𝟏𝟗𝟗𝟏𝟏°

d. What is the measure of ∠𝑪𝑪𝑬𝑬𝑮𝑮? Identify the angle relationship used to get your answer.

∠𝑪𝑪𝑬𝑬𝑮𝑮 = ∠𝑪𝑪𝑬𝑬𝑬𝑬 + ∠𝑬𝑬𝑬𝑬𝑮𝑮

∠𝑪𝑪𝑬𝑬𝑮𝑮 = 𝟏𝟏𝟗𝟗𝟏𝟏+ 𝟐𝟐𝟕𝟕 ∠𝑪𝑪𝑬𝑬𝑮𝑮 = 𝟏𝟏𝟑𝟑𝟗𝟗

𝒎𝒎∠𝑪𝑪𝑬𝑬𝑮𝑮 = 𝟏𝟏𝟑𝟑𝟗𝟗°

To determine the measure of ∠𝑪𝑪𝑬𝑬𝑮𝑮, you need to add the measures of adjacent angles ∠𝑪𝑪𝑬𝑬𝑬𝑬 and ∠𝑬𝑬𝑬𝑬𝑮𝑮.


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Name ___________________________________________________ Date____________________


Exit Ticket Write an equation for the angle relationship shown in the figure and solve for 𝑥𝑥. Find the measures of ∠𝑅𝑅𝑅𝑅𝑅𝑅 and ∠𝑇𝑇𝑅𝑅𝑇𝑇.

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Write an equation for the angle relationship shown in the figure and solve for 𝒙𝒙. Find the measures of ∠𝑹𝑹𝑹𝑹𝑹𝑹 and ∠𝑻𝑻𝑹𝑹𝑻𝑻.

𝟑𝟑𝒙𝒙 + 𝟗𝟗𝟗𝟗 + 𝟏𝟏𝒙𝒙 + 𝟐𝟐𝟐𝟐𝟏𝟏 = 𝟑𝟑𝟕𝟕𝟗𝟗 𝟕𝟕𝒙𝒙 + 𝟑𝟑𝟏𝟏𝟏𝟏 = 𝟑𝟑𝟕𝟕𝟗𝟗

𝟕𝟕𝒙𝒙 + 𝟑𝟑𝟏𝟏𝟏𝟏 − 𝟑𝟑𝟏𝟏𝟏𝟏 = 𝟑𝟑𝟕𝟕𝟗𝟗 − 𝟑𝟑𝟏𝟏𝟏𝟏 𝟕𝟕𝒙𝒙 = 𝟏𝟏𝟗𝟗

�𝟏𝟏𝟕𝟕�𝟕𝟕𝒙𝒙 = �

𝟏𝟏𝟕𝟕�𝟏𝟏𝟗𝟗

𝒙𝒙 = 𝟕𝟕

𝒎𝒎∠𝑹𝑹𝑹𝑹𝑹𝑹 = 𝟑𝟑(𝟕𝟕°) = 𝟐𝟐𝟏𝟏°

𝒎𝒎∠𝑻𝑻𝑹𝑹𝑻𝑻 = 𝟏𝟏(𝟕𝟕°) = 𝟐𝟐𝟏𝟏°


In a complete sentence, describe the angle relationships in each diagram. Write an equation for the angle relationship(s) shown in the figure, and solve for the indicated unknown angle. You can check your answers by measuring each angle with a protractor.

1. Find the measures of ∠𝑬𝑬𝑪𝑪𝑬𝑬, ∠𝑪𝑪𝑪𝑪𝑬𝑬, and ∠𝑪𝑪𝑪𝑪𝑪𝑪.

∠𝑮𝑮𝑪𝑪𝑬𝑬, ∠𝑬𝑬𝑪𝑪𝑬𝑬, ∠𝑪𝑪𝑪𝑪𝑬𝑬, and ∠𝑪𝑪𝑪𝑪𝑪𝑪 are angles on a line and their measures have a sum of 𝟏𝟏𝟏𝟏𝟗𝟗°.

𝟕𝟕𝒙𝒙 + 𝟏𝟏𝒙𝒙 + 𝟐𝟐𝒙𝒙 + 𝟑𝟑𝟗𝟗 = 𝟏𝟏𝟏𝟏𝟗𝟗 𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟑𝟑𝟗𝟗 = 𝟏𝟏𝟏𝟏𝟗𝟗

𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟑𝟑𝟗𝟗 − 𝟑𝟑𝟗𝟗 = 𝟏𝟏𝟏𝟏𝟗𝟗 − 𝟑𝟑𝟗𝟗 𝟏𝟏𝟐𝟐𝒙𝒙 = 𝟏𝟏𝟓𝟓𝟗𝟗

𝒙𝒙 = 𝟏𝟏𝟐𝟐.𝟓𝟓

𝒎𝒎∠𝑬𝑬𝑪𝑪𝑬𝑬 = 𝟐𝟐(𝟏𝟏𝟐𝟐.𝟓𝟓°) = 𝟐𝟐𝟓𝟓°

𝒎𝒎∠𝑪𝑪𝑪𝑪𝑬𝑬 = 𝟏𝟏(𝟏𝟏𝟐𝟐.𝟓𝟓°) = 𝟓𝟓𝟗𝟗°

𝒎𝒎∠𝑪𝑪𝑪𝑪𝑪𝑪 = 𝟕𝟕(𝟏𝟏𝟐𝟐.𝟓𝟓°) = 𝟕𝟕𝟓𝟓°

2. Find the measure of 𝒂𝒂.

Angles 𝒂𝒂°, 𝟐𝟐𝟕𝟕°, 𝒂𝒂°, and 𝟏𝟏𝟐𝟐𝟕𝟕° are angles at a point and have a sum of 𝟑𝟑𝟕𝟕𝟗𝟗°.

𝒂𝒂 + 𝟏𝟏𝟐𝟐𝟕𝟕+ 𝒂𝒂 + 𝟐𝟐𝟕𝟕 = 𝟑𝟑𝟕𝟕𝟗𝟗

𝟐𝟐𝒂𝒂 + 𝟏𝟏𝟓𝟓𝟐𝟐 = 𝟑𝟑𝟕𝟕𝟗𝟗 𝟐𝟐𝒂𝒂 + 𝟏𝟏𝟓𝟓𝟐𝟐 − 𝟏𝟏𝟓𝟓𝟐𝟐 = 𝟑𝟑𝟕𝟕𝟗𝟗 − 𝟏𝟏𝟓𝟓𝟐𝟐

𝟐𝟐𝒂𝒂 = 𝟐𝟐𝟗𝟗𝟏𝟏

�𝟏𝟏𝟐𝟐�𝟐𝟐𝒂𝒂 = �

𝟏𝟏𝟐𝟐�𝟐𝟐𝟗𝟗𝟏𝟏

𝒂𝒂 = 𝟏𝟏𝟗𝟗𝟏𝟏

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3. Find the measures of 𝒙𝒙 and 𝒚𝒚.

Angles 𝒚𝒚° and 𝟕𝟕𝟓𝟓° and angles 𝟐𝟐𝟓𝟓° and 𝒙𝒙° have a sum of 𝟗𝟗𝟗𝟗°.

𝒙𝒙 + 𝟐𝟐𝟓𝟓 = 𝟗𝟗𝟗𝟗

𝒙𝒙 + 𝟐𝟐𝟓𝟓 − 𝟐𝟐𝟓𝟓 = 𝟗𝟗𝟗𝟗 − 𝟐𝟐𝟓𝟓 𝒙𝒙 = 𝟕𝟕𝟓𝟓

𝟕𝟕𝟓𝟓 + 𝒚𝒚 = 𝟗𝟗𝟗𝟗 𝟕𝟕𝟓𝟓 + 𝒚𝒚 = 𝟗𝟗𝟗𝟗

𝟕𝟕𝟓𝟓 − 𝟕𝟕𝟓𝟓 + 𝒚𝒚 = 𝟗𝟗𝟗𝟗 − 𝟕𝟕𝟓𝟓 𝒚𝒚 = 𝟐𝟐𝟓𝟓

4. Find the measure of ∠𝑱𝑱𝑪𝑪𝑱𝑱.

Adjacent angles 𝒙𝒙° and 𝟏𝟏𝟓𝟓° together are vertically opposite from and are equal to angle 𝟏𝟏𝟏𝟏°.

𝒙𝒙 + 𝟏𝟏𝟓𝟓 = 𝟏𝟏𝟏𝟏 𝒙𝒙 + 𝟏𝟏𝟓𝟓 − 𝟏𝟏𝟓𝟓 = 𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟓𝟓

𝒙𝒙 = 𝟕𝟕𝟕𝟕

𝒎𝒎∠𝑱𝑱𝑪𝑪𝑱𝑱 = 𝟕𝟕𝟕𝟕°

5. Find the measures of ∠𝑱𝑱𝑪𝑪𝑬𝑬 and ∠𝑪𝑪𝑪𝑪𝑬𝑬.

The measures of adjacent angles ∠𝑪𝑪𝑪𝑪𝑬𝑬 and ∠𝑱𝑱𝑪𝑪𝑬𝑬 have a sum of the measure of ∠𝑪𝑪𝑪𝑪𝑱𝑱, which is vertically opposite from and equal to the measurement of ∠𝑪𝑪𝑪𝑪𝑬𝑬.

𝟐𝟐𝒙𝒙 + 𝟑𝟑𝒙𝒙 + 𝟕𝟕𝟗𝟗 = 𝟏𝟏𝟏𝟏𝟗𝟗

𝟓𝟓𝒙𝒙 = 𝟏𝟏𝟏𝟏𝟗𝟗

�𝟏𝟏𝟓𝟓�𝟓𝟓𝒙𝒙 = �

𝟏𝟏𝟓𝟓�𝟏𝟏𝟏𝟏𝟗𝟗

𝒙𝒙 = 𝟐𝟐𝟐𝟐

𝒎𝒎∠𝑱𝑱𝑪𝑪𝑬𝑬 = 𝟑𝟑(𝟐𝟐𝟐𝟐°) = 𝟕𝟕𝟕𝟕°

𝒎𝒎∠𝑪𝑪𝑪𝑪𝑬𝑬 = 𝟐𝟐(𝟐𝟐𝟐𝟐°) = 𝟏𝟏𝟏𝟏°

6. The measure of ∠𝑹𝑹𝑬𝑬𝑻𝑻 is 𝒃𝒃°. The measure of ∠𝑻𝑻𝑬𝑬𝑹𝑹 is five more than two times ∠𝑹𝑹𝑬𝑬𝑻𝑻. The measure of ∠𝑹𝑹𝑬𝑬𝑹𝑹 is twelve less than eight times the measure of ∠𝑹𝑹𝑬𝑬𝑻𝑻. Find the measures of ∠𝑹𝑹𝑬𝑬𝑻𝑻, ∠𝑻𝑻𝑬𝑬𝑹𝑹, and ∠𝑹𝑹𝑬𝑬𝑹𝑹.

∠𝑹𝑹𝑬𝑬𝑹𝑹, ∠𝑹𝑹𝑬𝑬𝑻𝑻, and ∠𝑻𝑻𝑬𝑬𝑹𝑹 are angles on a line and their measures have a sum of 𝟏𝟏𝟏𝟏𝟗𝟗°.

(𝟏𝟏𝒃𝒃 − 𝟏𝟏𝟐𝟐) + 𝒃𝒃 + (𝟐𝟐𝒃𝒃 + 𝟓𝟓) = 𝟏𝟏𝟏𝟏𝟗𝟗 𝟏𝟏𝟏𝟏𝒃𝒃 − 𝟕𝟕 = 𝟏𝟏𝟏𝟏𝟗𝟗

𝟏𝟏𝟏𝟏𝒃𝒃 − 𝟕𝟕 + 𝟕𝟕 = 𝟏𝟏𝟏𝟏𝟗𝟗+ 𝟕𝟕 𝟏𝟏𝟏𝟏𝒃𝒃 = 𝟏𝟏𝟏𝟏𝟕𝟕

�𝟏𝟏𝟏𝟏𝟏𝟏�𝟏𝟏𝟏𝟏𝒃𝒃 = �

𝟏𝟏𝟏𝟏𝟏𝟏�𝟏𝟏𝟏𝟏𝟕𝟕

𝒃𝒃 = 𝟏𝟏𝟕𝟕

𝒎𝒎∠𝑹𝑹𝑬𝑬𝑻𝑻 = (𝟏𝟏𝟕𝟕°) = 𝟏𝟏𝟕𝟕°

𝒎𝒎∠𝑻𝑻𝑬𝑬𝑹𝑹 = 𝟐𝟐(𝟏𝟏𝟕𝟕°) + 𝟓𝟓° = 𝟑𝟑𝟗𝟗°

𝒎𝒎∠𝑹𝑹𝑬𝑬𝑹𝑹 = 𝟏𝟏(𝟏𝟏𝟕𝟕°) − 𝟏𝟏𝟐𝟐° = 𝟏𝟏𝟐𝟐𝟏𝟏°

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7. Find the measures of ∠𝑱𝑱𝑹𝑹𝑬𝑬 and ∠𝑪𝑪𝑹𝑹𝑮𝑮.

∠𝑪𝑪𝑹𝑹𝑮𝑮, ∠𝑪𝑪𝑹𝑹𝑱𝑱, and ∠𝑱𝑱𝑹𝑹𝑬𝑬 are adjacent angles whose measures have a sum of 𝟗𝟗𝟗𝟗°.

𝟐𝟐𝒚𝒚+ 𝟐𝟐𝟏𝟏 + 𝒚𝒚 = 𝟗𝟗𝟗𝟗 𝟑𝟑𝒚𝒚 + 𝟐𝟐𝟏𝟏 = 𝟗𝟗𝟗𝟗

𝟑𝟑𝒚𝒚 + 𝟐𝟐𝟏𝟏 − 𝟐𝟐𝟏𝟏 = 𝟗𝟗𝟗𝟗 − 𝟐𝟐𝟏𝟏 𝟑𝟑𝒚𝒚 = 𝟕𝟕𝟗𝟗

�𝟏𝟏𝟑𝟑�𝟑𝟑𝒚𝒚 = �

𝟏𝟏𝟑𝟑�𝟕𝟕𝟗𝟗

𝒚𝒚 = 𝟐𝟐𝟑𝟑

𝒎𝒎∠𝑱𝑱𝑹𝑹𝑬𝑬 = 𝟐𝟐(𝟐𝟐𝟑𝟑°) = 𝟏𝟏𝟕𝟕°

𝒎𝒎∠𝑪𝑪𝑹𝑹𝑮𝑮 = (𝟐𝟐𝟑𝟑°) = 𝟐𝟐𝟑𝟑°

8. The measures of three angles at a point are in the ratio of 𝟐𝟐 ∶ 𝟑𝟑 ∶ 𝟓𝟓. Find the measures of the angles.

∠𝑪𝑪 = 𝟐𝟐𝒙𝒙, ∠𝑬𝑬 = 𝟑𝟑𝒙𝒙, ∠𝑪𝑪 = 𝟓𝟓𝒙𝒙

𝟐𝟐𝒙𝒙 + 𝟑𝟑𝒙𝒙 + 𝟓𝟓𝒙𝒙 = 𝟑𝟑𝟕𝟕𝟗𝟗

𝟏𝟏𝟗𝟗𝒙𝒙 = 𝟑𝟑𝟕𝟕𝟗𝟗

�𝟏𝟏𝟏𝟏𝟗𝟗�𝟏𝟏𝟗𝟗𝒙𝒙 = �

𝟏𝟏𝟏𝟏𝟗𝟗�𝟑𝟑𝟕𝟕𝟗𝟗

𝒙𝒙 = 𝟑𝟑𝟕𝟕

∠𝑪𝑪 = 𝟐𝟐(𝟑𝟑𝟕𝟕°) = 𝟕𝟕𝟐𝟐°

∠𝑬𝑬 = 𝟑𝟑(𝟑𝟑𝟕𝟕°) = 𝟏𝟏𝟗𝟗𝟏𝟏°

∠𝑪𝑪 = 𝟓𝟓(𝟑𝟑𝟕𝟕°) = 𝟏𝟏𝟏𝟏𝟗𝟗°

9. The sum of the measures of two adjacent angles is 𝟕𝟕𝟐𝟐°. The ratio of the smaller angle to the larger angle is 𝟏𝟏 ∶ 𝟑𝟑. Find the measures of each angle.

∠𝑪𝑪 = 𝒙𝒙, ∠𝑬𝑬 = 𝟑𝟑𝒙𝒙

𝒙𝒙 + 𝟑𝟑𝒙𝒙 = 𝟕𝟕𝟐𝟐 𝟏𝟏𝒙𝒙 = 𝟕𝟕𝟐𝟐

�𝟏𝟏𝟏𝟏� (𝟏𝟏𝒙𝒙) = �

𝟏𝟏𝟏𝟏� (𝟕𝟕𝟐𝟐)

𝒙𝒙 = 𝟏𝟏𝟏𝟏

∠𝑪𝑪 = (𝟏𝟏𝟏𝟏°) = 𝟏𝟏𝟏𝟏°

∠𝑬𝑬 = 𝟑𝟑(𝟏𝟏𝟏𝟏°) = 𝟓𝟓𝟏𝟏°

10. Find the measures of ∠𝑪𝑪𝑹𝑹𝑪𝑪 and ∠𝑬𝑬𝑹𝑹𝑬𝑬.

𝟏𝟏𝒙𝒙 + 𝟓𝟓𝒙𝒙 = 𝟏𝟏𝟗𝟗𝟏𝟏 𝟗𝟗𝒙𝒙 = 𝟏𝟏𝟗𝟗𝟏𝟏

�𝟏𝟏𝟗𝟗�𝟗𝟗𝒙𝒙 = �

𝟏𝟏𝟗𝟗�𝟏𝟏𝟗𝟗𝟏𝟏

𝒙𝒙 = 𝟏𝟏𝟐𝟐

𝒎𝒎∠𝑪𝑪𝑹𝑹𝑪𝑪 = 𝟓𝟓(𝟏𝟏𝟐𝟐°) = 𝟕𝟕𝟗𝟗°

𝒎𝒎∠𝑬𝑬𝑹𝑹𝑬𝑬 = 𝟏𝟏(𝟏𝟏𝟐𝟐°) = 𝟏𝟏𝟏𝟏°

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Lesson 12: Properties of Inequalities

7•3Lesson 12


Student Outcomes

Students justify the properties of inequalities that are denoted by < (less than), ≤ (less than or equal to), > (greater than), and ≥ (greater than or equal to).

Classwork

Rapid Whiteboard Exchange (10 minutes): Equations

Students complete a rapid whiteboard exchange where they practice their knowledge of solving linear equations in the form 𝑝𝑝𝑝𝑝 + 𝑞𝑞 = 𝑟𝑟 and 𝑝𝑝(𝑝𝑝 + 𝑞𝑞) = 𝑟𝑟.


Review the descriptions of preserves the inequality symbol and reverses the inequality symbol with students.

Example 1

Preserves the inequality symbol: means the inequality symbol stays the same.

Reverses the inequality symbol: means the inequality symbol switches less than with greater than and less than or equal to with greater than or equal to.

Exploratory Challenge (20 minutes)

Split students into four groups. Discuss the directions.

There are four stations. Provide each station with two cubes containing integers. (Cube templates provided at the end of the document.) At each station, students record their results in their student materials. (An example is provided for each station.)

1. Roll each die, recording the numbers under the first and third columns. Write an inequality symbol that makes the statement true. Repeat this four times to complete the four rows in the table.

2. Perform the operation indicated at the station (adding or subtracting a number, writing opposites, multiplying or dividing by a number), and write a new inequality statement.

3. Determine if the inequality symbol is preserved or reversed when the operation is performed.

4. Rotate to a new station after five minutes.

MP.2 &

MP.4

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Station 1: Add or Subtract a Number to Both Sides of the Inequality

Station 1

Die 1 Inequality Die 2 Operation New Inequality Inequality Symbol Preserved or Reversed?

−𝟑𝟑 < 𝟓𝟓 Add 𝟐𝟐 −𝟑𝟑 + 𝟐𝟐 < 𝟓𝟓 + 𝟐𝟐 −𝟏𝟏 < 𝟕𝟕

Preserved

Add −𝟑𝟑

Subtract 𝟐𝟐

Subtract −𝟏𝟏

Add 𝟏𝟏

Examine the results. Make a statement about what you notice, and justify it with evidence.

When a number is added or subtracted to both numbers being compared, the symbol stays the same, and the inequality symbol is preserved.

Station 2: Multiply each term by −1

Station 2


−𝟑𝟑 < 𝟒𝟒 Multiply by

−𝟏𝟏 (−𝟏𝟏)(−𝟑𝟑) < (−𝟏𝟏)(𝟒𝟒)

𝟑𝟑 > −𝟒𝟒 Reversed

Multiply by −𝟏𝟏




Examine the results. Make a statement about what you notice and justify it with evidence.

When both numbers are multiplied by −𝟏𝟏, the symbol changes, and the inequality symbol is reversed.

Scaffolding:

Guide students in writing a statement using the following:

When −1 is multiplied to both numbers, the symbol _________; therefore, the inequality symbol is ___________.

Scaffolding:


When a number is added or subtracted to both numbers being compared, the symbol ____________; therefore, the inequality symbol is __________.

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7•3 Lesson 12

Station 3: Multiply or Divide Both Sides of the Inequality by a Positive Number

Station 3


−𝟐𝟐 > −𝟒𝟒 Multiply by 𝟏𝟏𝟐𝟐 (−𝟐𝟐) �

𝟏𝟏𝟐𝟐� > (−𝟒𝟒) �

𝟏𝟏𝟐𝟐�

−𝟏𝟏 > −𝟐𝟐 Preserved

Multiply by 𝟐𝟐

Divide by 𝟐𝟐

Divide by 𝟏𝟏𝟐𝟐

Multiply by 𝟑𝟑

Examine the results. Make a statement about what you notice, and justify it with evidence.

When both numbers being compared are multiplied by or divided by a positive number, the symbol stays the same, and the inequality symbol is preserved.

Station 4: Multiply or Divide Both Sides of the Inequality by a Negative Number

Station 4


𝟑𝟑 > −𝟐𝟐 Multiply by

−𝟐𝟐 𝟑𝟑(−𝟐𝟐) > (−𝟐𝟐)(−𝟐𝟐)

−𝟔𝟔 < 𝟒𝟒 Reversed

Multiply by −𝟑𝟑

Divide by −𝟐𝟐

Divide by

−𝟏𝟏𝟐𝟐

Multiply by

−𝟏𝟏𝟐𝟐

Examine the results. Make a statement about what you notice and justify it with evidence.

When both numbers being compared are multiplied by or divided by a negative number, the symbol changes, and the inequality symbol is reversed.

Scaffolding:


When both numbers being compared are multiplied by or divided by a positive number, the symbol _______________; therefore, the inequality symbol is __________.

Scaffolding: Guide students in writing a statement using the following:

When both numbers being compared are multiplied by or divided by a negative number, the symbol ___________; therefore, the inequality symbol is _________.

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7•3 Lesson 12

Discussion

Summarize the findings and complete the Lesson Summary in the student materials.

To summarize, when does an inequality change (reverse), and when does it stay the same (preserve)? The inequality reverses when we multiply or divide the expressions on both sides of the inequality by a

negative number. The inequality stays the same for all other cases.

Exercise (5 minutes) Exercise

Complete the following chart using the given inequality, and determine an operation in which the inequality symbol is preserved and an operation in which the inequality symbol is reversed. Explain why this occurs.

Solutions may vary. A sample student response is below.

Inequality Operation and New Inequality Which Preserves the Inequality

Symbol

Operation and New Inequality Which Reverses

the Inequality Symbol Explanation

𝟐𝟐 < 𝟓𝟓

Add 𝟒𝟒 to both sides. 𝟐𝟐 < 𝟓𝟓

𝟐𝟐 + 𝟒𝟒 < 𝟓𝟓+ 𝟒𝟒 𝟔𝟔 < 𝟗𝟗

Multiply both sides by −𝟒𝟒. −𝟖𝟖 > −𝟐𝟐𝟐𝟐

Adding a number to both sides of an inequality preserves the inequality symbol. Multiplying both sides of an inequality by a negative number reverses the inequality symbol.

−𝟒𝟒 > −𝟔𝟔

Subtract 𝟑𝟑 from both sides. −𝟒𝟒 > −𝟔𝟔

−𝟒𝟒− 𝟑𝟑 > −𝟔𝟔− 𝟑𝟑 −𝟕𝟕 > −𝟗𝟗

Divide both sides by −𝟐𝟐. 𝟐𝟐 < 𝟑𝟑

Subtracting a number from both sides of an inequality preserves the inequality symbol. Dividing both sides of an inequality by a negative number reverses the inequality symbol.

−𝟏𝟏 ≤ 𝟐𝟐

Multiply both sides by 𝟑𝟑. −𝟏𝟏 ≤ 𝟐𝟐

−𝟏𝟏(𝟑𝟑) ≤ 𝟐𝟐(𝟑𝟑) −𝟑𝟑 ≤ 𝟔𝟔

Multiply both sides by −𝟏𝟏. 𝟏𝟏 ≥ −𝟐𝟐

Multiplying both sides of an inequality by a positive number preserves the inequality symbol. Multiplying both sides of an inequality by a negative number reverses the inequality symbol.

−𝟐𝟐+ (−𝟑𝟑)< −𝟑𝟑− 𝟏𝟏

Add 𝟓𝟓 to both sides. −𝟐𝟐 + (−𝟑𝟑) < −𝟑𝟑 − 𝟏𝟏

−𝟐𝟐+ (−𝟑𝟑) + 𝟓𝟓 < −𝟑𝟑 − 𝟏𝟏 + 𝟓𝟓 𝟐𝟐 < 𝟏𝟏

Multiply each side by−𝟏𝟏𝟐𝟐.

−𝟐𝟐 + (−𝟑𝟑) < −𝟑𝟑− 𝟏𝟏 −𝟓𝟓 < −𝟒𝟒

�−𝟏𝟏𝟐𝟐� (−𝟓𝟓) > �−

𝟏𝟏𝟐𝟐� (−𝟒𝟒)

𝟓𝟓𝟐𝟐

> 𝟐𝟐

Adding a number to both sides of an inequality preserves the inequality symbol. Multiplying both sides of an inequality by a negative number reverses the inequality symbol.

Closing (3 minutes)

What does it mean for an inequality to be preserved? What does it mean for the inequality to be reversed?

When an operation is done to both sides and the inequality does not change, it is preserved. If the inequality does change, it is reversed. For example, less than would become greater than.

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When does a greater than become a less than?

When both sides are multiplied or divided by a negative, the inequality is reversed.


Lesson Summary

When both sides of an inequality are added or subtracted by a number, the inequality symbol stays the same, and the inequality symbol is said to be preserved.

When both sides of an inequality are multiplied or divided by a positive number, the inequality symbol stays the same, and the inequality symbol is said to be preserved.

When both sides of an inequality are multiplied or divided by a negative number, the inequality symbol switches from < to > or from > to <. The inequality symbol is reversed.

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Name ___________________________________________________ Date____________________


Exit Ticket 1. Given the initial inequality −4 < 7, state possible values for 𝑐𝑐 that would satisfy the following inequalities.

a. 𝑐𝑐(−4) < 𝑐𝑐(7)

b. 𝑐𝑐(−4) > 𝑐𝑐(7)

c. 𝑐𝑐(−4) = 𝑐𝑐(7)

2. Given the initial inequality 2 > −4, identify which operation preserves the inequality symbol and which operation reverses the inequality symbol. Write the new inequality after the operation is performed.

a. Multiply both sides by −2.

b. Add −2 to both sides.

c. Divide both sides by 2.

d. Multiply both sides by − 12.

e. Subtract −3 from both sides.

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7•3 Lesson 12


1. Given the initial inequality −𝟒𝟒 < 𝟕𝟕, state possible values for 𝒄𝒄 that would satisfy the following inequalities.

a. 𝒄𝒄(−𝟒𝟒) < 𝒄𝒄(𝟕𝟕)

𝒄𝒄 > 𝟐𝟐

b. 𝒄𝒄(−𝟒𝟒) > 𝒄𝒄(𝟕𝟕)

𝒄𝒄 < 𝟐𝟐

c. 𝒄𝒄(−𝟒𝟒) = 𝒄𝒄(𝟕𝟕)

𝒄𝒄 = 𝟐𝟐

2. Given the initial inequality 𝟐𝟐 > −𝟒𝟒, identify which operation preserves the inequality symbol and which operation reverses the inequality symbol. Write the new inequality after the operation is performed.

a. Multiply both sides by −𝟐𝟐.

The inequality symbol is reversed.

𝟐𝟐 > −𝟒𝟒 𝟐𝟐(−𝟐𝟐) < −𝟒𝟒(−𝟐𝟐)

−𝟒𝟒 < 𝟖𝟖

b. Add −𝟐𝟐 to both sides.

The inequality symbol is preserved.

𝟐𝟐 > −𝟒𝟒

𝟐𝟐 + (−𝟐𝟐) > −𝟒𝟒 + (−𝟐𝟐) 𝟐𝟐 > −𝟔𝟔

c. Divide both sides by 𝟐𝟐.


𝟐𝟐 > −𝟒𝟒

𝟐𝟐 ÷ 𝟐𝟐 > −𝟒𝟒 ÷ 𝟐𝟐 𝟏𝟏 > −𝟐𝟐

d. Multiply both sides by −𝟏𝟏𝟐𝟐.

Inequality symbol is reversed.

𝟐𝟐 > −𝟒𝟒

𝟐𝟐�−𝟏𝟏𝟐𝟐� < −𝟒𝟒�−

𝟏𝟏𝟐𝟐�

−𝟏𝟏 < 𝟐𝟐

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7•3 Lesson 12

e. Subtract −𝟑𝟑 from both sides.


𝟐𝟐 > −𝟒𝟒

𝟐𝟐 − (−𝟑𝟑) > −𝟒𝟒 − (−𝟑𝟑) 𝟓𝟓 > −𝟏𝟏


1. For each problem, use the properties of inequalities to write a true inequality statement. The two integers are −𝟐𝟐 and −𝟓𝟓.

a. Write a true inequality statement.

−𝟓𝟓 < −𝟐𝟐

b. Subtract −𝟐𝟐 from each side of the inequality. Write a true inequality statement.

−𝟕𝟕 < −𝟒𝟒

c. Multiply each number by −𝟑𝟑. Write a true inequality statement.

𝟏𝟏𝟓𝟓 > 𝟔𝟔

2. On a recent vacation to the Caribbean, Kay and Tony wanted to explore the ocean elements. One day they went in a submarine 𝟏𝟏𝟓𝟓𝟐𝟐 feet below sea level. The second day they went scuba diving 𝟕𝟕𝟓𝟓 feet below sea level.

a. Write an inequality comparing the submarine’s elevation and the scuba diving elevation.

−𝟏𝟏𝟓𝟓𝟐𝟐 < −𝟕𝟕𝟓𝟓

b. If they only were able to go one-fifth of the capable elevations, write a new inequality to show the elevations they actually achieved.

−𝟑𝟑𝟐𝟐 < −𝟏𝟏𝟓𝟓

c. Was the inequality symbol preserved or reversed? Explain.

The inequality symbol was preserved because the number that was multiplied to both sides was NOT negative.

3. If 𝒂𝒂 is a negative integer, then which of the number sentences below is true? If the number sentence is not true, give a reason.

a. 𝟓𝟓 + 𝒂𝒂 < 𝟓𝟓

True

b. 𝟓𝟓 + 𝒂𝒂 > 𝟓𝟓

False because adding a negative number to 𝟓𝟓 will decrease 𝟓𝟓, which will not be greater than 𝟓𝟓.

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7•3 Lesson 12

c. 𝟓𝟓 − 𝒂𝒂 > 𝟓𝟓

True

d. 𝟓𝟓 − 𝒂𝒂 < 𝟓𝟓

False because subtracting a negative number is adding a positive number to 𝟓𝟓, which will be larger than 𝟓𝟓.

e. 𝟓𝟓𝒂𝒂 < 𝟓𝟓

True

f. 𝟓𝟓𝒂𝒂 > 𝟓𝟓

False because a negative number multiplied by a positive number is negative, which will be less than 𝟓𝟓.

g. 𝟓𝟓 + 𝒂𝒂 > 𝒂𝒂

True

h. 𝟓𝟓 + 𝒂𝒂 < 𝒂𝒂

False because adding 𝟓𝟓 to a negative number is greater than the negative number itself.

i. 𝟓𝟓 − 𝒂𝒂 > 𝒂𝒂

True

j. 𝟓𝟓 − 𝒂𝒂 < 𝒂𝒂

False because subtracting a negative number is the same as adding a positive number, which is greater than the negative number itself.

k. 𝟓𝟓𝒂𝒂 > 𝒂𝒂

False because a negative number multiplied by a 𝟓𝟓 is negative and will be 𝟓𝟓 times smaller than 𝒂𝒂.

l. 𝟓𝟓𝒂𝒂 < 𝒂𝒂

True

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7•3 Lesson 12

Equations

Progression of Exercises

Determine the value of the variable.

Set 1

1. 𝑝𝑝 + 1 = 5 𝒙𝒙 = 𝟒𝟒

2. 𝑝𝑝 + 3 = 5 𝒙𝒙 = 𝟐𝟐

3. 𝑝𝑝 + 6 = 5 𝒙𝒙 = −𝟏𝟏

4. 𝑝𝑝 − 5 = 2 𝒙𝒙 = 𝟕𝟕

5. 𝑝𝑝 − 5 = 8

𝒙𝒙 = 𝟏𝟏𝟑𝟑

Set 2

1. 3𝑝𝑝 = 15 𝒙𝒙 = 𝟓𝟓

2. 3𝑝𝑝 = 0 𝒙𝒙 = 𝟐𝟐

3. 3𝑝𝑝 = −3 𝒙𝒙 = −𝟏𝟏

4. −9𝑝𝑝 = 18 𝒙𝒙 = −𝟐𝟐

5. −𝑝𝑝 = 18 𝒙𝒙 = −𝟏𝟏𝟖𝟖

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Set 3

1. 17𝑝𝑝 = 5

𝒙𝒙 = 𝟑𝟑𝟓𝟓

2. 27𝑝𝑝 = 10

𝒙𝒙 = 𝟑𝟑𝟓𝟓

3. 37𝑝𝑝 = 15

𝒙𝒙 = 𝟑𝟑𝟓𝟓

4. 47𝑝𝑝 = 20

𝒙𝒙 = 𝟑𝟑𝟓𝟓

5. −57𝑝𝑝 = −25

𝒙𝒙 = 𝟑𝟑𝟓𝟓

Set 4

1. 2𝑝𝑝 + 4 = 12 𝒙𝒙 = 𝟒𝟒

2. 2𝑝𝑝 − 5 = 13 𝒙𝒙 = 𝟗𝟗

3. 2𝑝𝑝 + 6 = 14 𝒙𝒙 = 𝟒𝟒

4. 3𝑝𝑝 − 6 = 18 𝒙𝒙 = 𝟖𝟖

5. −4𝑝𝑝 + 6 = 22 𝒙𝒙 = −𝟒𝟒

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Set 5

1. 2𝑝𝑝 + 0.5 = 6.5 𝒙𝒙 = 𝟑𝟑

2. 3𝑝𝑝 − 0.5 = 8.5 𝒙𝒙 = 𝟑𝟑

3. 5𝑝𝑝 + 3 = 8.5 𝒙𝒙 = 𝟏𝟏.𝟏𝟏

4. 5𝑝𝑝 − 4 = 1.5 𝒙𝒙 = 𝟏𝟏.𝟏𝟏

5. −7𝑝𝑝 + 1.5 = 5 𝒙𝒙 = −𝟐𝟐.𝟓𝟓

Set 6

1. 2(𝑝𝑝 + 3) = 4 𝒙𝒙 = −𝟏𝟏

2. 5(𝑝𝑝 + 3) = 10 𝒙𝒙 = −𝟏𝟏

3. 5(𝑝𝑝 − 3) = 10 𝒙𝒙 = 𝟓𝟓

4. −2(𝑝𝑝 − 3) = 8

𝒙𝒙 = −𝟏𝟏

5. −3(𝑝𝑝 + 4) = 3 𝒙𝒙 = −𝟓𝟓

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7•3 Lesson 12

Die Templates

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Lesson 13: Inequalities

7•3 Lesson 13


Student Outcomes

Students understand that an inequality is a statement that one expression is less than (or equal to) or greater than (or equal to) another expression, such as 2𝑥𝑥 + 3 < 5 or 3𝑥𝑥 + 50 ≥ 100.

Students interpret a solution to an inequality as a number that makes the inequality true when substituted for the variable.

Students convert arithmetic inequalities into a new inequality with variables (e.g., 2 × 6 + 3 > 12 to 2𝑚𝑚 + 3 > 12) and give a solution, such as 𝑚𝑚 = 6, to the new inequality. They check to see if different values of the variable make an inequality true or false.

Lesson Notes This lesson reviews the conceptual understanding of inequalities and introduces the “why” and “how” of moving from numerical expressions to algebraic inequalities.

Classwork

Opening Exercise (12 minutes): Writing Inequality Statements Opening Exercise: Writing Inequality Statements

Tarik is trying to save $𝟐𝟐𝟐𝟐𝟐𝟐.𝟒𝟒𝟒𝟒 to buy a new tablet. Right now, he has $𝟒𝟒𝟒𝟒 and can save $𝟑𝟑𝟑𝟑 a week from his allowance.

Write and evaluate an expression to represent the amount of money saved after …

2 weeks 𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟑𝟑(𝟐𝟐)

𝟒𝟒𝟒𝟒 + 𝟕𝟕𝟐𝟐

𝟏𝟏𝟏𝟏𝟐𝟐

3 weeks 𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟑𝟑(𝟑𝟑)

𝟒𝟒𝟒𝟒 + 𝟏𝟏𝟏𝟏𝟒𝟒

𝟏𝟏𝟐𝟐𝟒𝟒

4 weeks 𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟑𝟑(𝟒𝟒)

𝟒𝟒𝟒𝟒 + 𝟏𝟏𝟐𝟐𝟐𝟐

𝟏𝟏𝟒𝟒𝟐𝟐

5 weeks 𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟑𝟑(𝟐𝟐)

𝟒𝟒𝟒𝟒 + 𝟏𝟏𝟒𝟒𝟒𝟒

𝟐𝟐𝟑𝟑𝟒𝟒

6 weeks 𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟑𝟑(𝟐𝟐)

𝟒𝟒𝟒𝟒 + 𝟐𝟐𝟐𝟐𝟑𝟑

𝟐𝟐𝟐𝟐𝟑𝟑

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7•3 Lesson 13

7 weeks 𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟑𝟑(𝟕𝟕)

𝟒𝟒𝟒𝟒 + 𝟐𝟐𝟐𝟐𝟐𝟐

𝟑𝟑𝟒𝟒𝟐𝟐

8 weeks 𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟑𝟑(𝟑𝟑)

𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟒𝟒𝟒𝟒

𝟑𝟑𝟒𝟒𝟒𝟒

When will Tarik have enough money to buy the tablet?

From 6 weeks and onward

Write an inequality that will generalize the problem.

𝟑𝟑𝟑𝟑𝟑𝟑 + 𝟒𝟒𝟒𝟒 ≥ 𝟐𝟐𝟐𝟐𝟐𝟐.𝟒𝟒𝟒𝟒 Where 𝟑𝟑 represents the number of weeks it will take to save the money.

Discussion

Why is it possible to have more than one solution?

It is possible because the minimum amount of money Tarik needs to buy the tablet is $265.49. He can save more money than that, but he cannot have less than that amount. As more time passes, he will have saved more money. Therefore, any amount of time from 6 weeks onward will ensure he has enough money to purchase the tablet.

So, the minimum amount of money Tarik needs is $265.49, and he could have more but certainly not less. What inequality would demonstrate this?

Greater than or equal to Examine each of the numerical expressions previously, and write an inequality showing the actual amount of

money saved compared to what is needed. Then, determine if each inequality written is true or false.

2 weeks: 116 ≥ 265.49 False

3 weeks: 154 ≥ 265.49 False

4 weeks: 192 ≥ 265.49 False

5 weeks: 230 ≥ 265.49 False 6 weeks: 268 ≥ 265.49 True

7 weeks: 306 ≥ 265.49 True

8 weeks: 344 ≥ 265.49 True

How can this problem be generalized?

Instead of asking what amount of money was saved after a specific amount of time, the question can be asked: How long will it take Tarik to save enough money to buy the tablet?

Write an inequality that would generalize this problem for money being saved for 𝑤𝑤 weeks. 38𝑤𝑤 + 40 ≥ 265.49

Interpret the meaning of the 38 in the inequality 38𝑤𝑤 + 40 ≥ 265.49. The 38 represents the amount of money saved each week. As the weeks increase, the amount of

money increases. The $40 was the initial amount of money saved, not the amount saved every week.

MP.7

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7•3 Lesson 13

Example 1 (13 minutes): Evaluating Inequalities—Finding a Solution

Example 1: Evaluating Inequalities—Finding a Solution

The sum of two consecutive odd integers is more than −𝟏𝟏𝟐𝟐. Write several true numerical inequality expressions.

𝟐𝟐 + 𝟕𝟕 > −𝟏𝟏𝟐𝟐 𝟑𝟑 + 𝟐𝟐 > −𝟏𝟏𝟐𝟐 𝟏𝟏 + 𝟑𝟑 > −𝟏𝟏𝟐𝟐 −𝟏𝟏 + 𝟏𝟏 > −𝟏𝟏𝟐𝟐 −𝟑𝟑+ −𝟏𝟏 > −𝟏𝟏𝟐𝟐

𝟏𝟏𝟐𝟐 > −𝟏𝟏𝟐𝟐 𝟑𝟑 > −𝟏𝟏𝟐𝟐 𝟒𝟒 > −𝟏𝟏𝟐𝟐 𝟒𝟒 > −𝟏𝟏𝟐𝟐 −𝟒𝟒 > −𝟏𝟏𝟐𝟐

The sum of two consecutive odd integers is more than −𝟏𝟏𝟐𝟐. What is the smallest value that will make this true?

a. Write an inequality that can be used to find the smallest value that will make the statement true.

𝒙𝒙: an integer

𝟐𝟐𝒙𝒙 + 𝟏𝟏: odd integer

𝟐𝟐𝒙𝒙 + 𝟑𝟑: next consecutive odd integer

𝟐𝟐𝒙𝒙 + 𝟏𝟏+ 𝟐𝟐𝒙𝒙 + 𝟑𝟑 > −𝟏𝟏𝟐𝟐

b. Use if-then moves to solve the inequality written in part (a). Identify where the 𝟒𝟒′s and 𝟏𝟏′s were made using the if-then moves.

𝟒𝟒𝒙𝒙 + 𝟒𝟒 > −𝟏𝟏𝟐𝟐 𝟒𝟒𝒙𝒙 + 𝟒𝟒 − 𝟒𝟒 > −𝟏𝟏𝟐𝟐 − 𝟒𝟒

𝟒𝟒𝒙𝒙 + 𝟒𝟒 > −𝟏𝟏𝟐𝟐

�𝟏𝟏𝟒𝟒� (𝟒𝟒𝒙𝒙) > �

𝟏𝟏𝟒𝟒� (−𝟏𝟏𝟐𝟐)

𝒙𝒙 > −𝟒𝟒

If 𝒂𝒂 > 𝒃𝒃, then 𝒂𝒂 − 𝟒𝟒 > 𝒃𝒃 − 𝟒𝟒.

𝟒𝟒 was the result.

If 𝒂𝒂 > 𝒃𝒃, then 𝒂𝒂 �𝟏𝟏𝟒𝟒� > 𝒃𝒃�𝟏𝟏𝟒𝟒�.

𝟏𝟏 was the result.

c. What is the smallest value that will make this true? To find the odd integer, substitute −𝟒𝟒 for 𝒙𝒙 in 𝟐𝟐𝒙𝒙 + 𝟏𝟏.

𝟐𝟐(−𝟒𝟒) + 𝟏𝟏

−𝟑𝟑+ 𝟏𝟏

−𝟕𝟕

The values that will solve the original inequality are all the odd integers greater than −𝟕𝟕. Therefore, the smallest values that will make this true are −𝟐𝟐 and −𝟑𝟑.

Questions to discuss leading to writing the inequality:

What is the difference between consecutive integers and consecutive even or odd integers? Consecutive even/odd integers increase or decrease by 2 units compared to consecutive integers that

increase by 1 unit.

What inequality symbol represents is more than? Why?

> represents is more than because a number that is more than another number is bigger than the original number.

Scaffolding:

To ensure that the integers will be odd and not even, the first odd integer is one unit greater than or less than an even integer. If 𝑥𝑥 is an integer, then 2𝑥𝑥 would ensure an even integer, and 2𝑥𝑥 + 1 would be an odd integer since it is one unit greater than an even integer.

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Questions leading to finding a solution:

What is a solution set of an inequality?

A solution set contains more than one number that makes the inequality a true statement.

Is −3 a solution to our inequality in part (a)?

Yes. When the value of −3 is substituted into the inequality, the resulting statement is true. Could −4 be a solution to our inequality in part (a)?

Substituting −4 does not result in a true statement because −12 is equal to, but not greater than −12.

We have found that 𝑥𝑥 = −3 is a solution to the inequality in part (a) where 𝑥𝑥 = −4 and 𝑥𝑥 = −5 are not. What is meant by the minimum value in this inequality? Explain.

The minimum value is the smallest value that makes the inequality true. −3 is not the minimum value

because there are rational numbers that are smaller than −3 but greater than −4. For example, −3 12

is smaller than −3 but still creates a true statement.

How is solving an inequality similar to solving an equation? How is it different?

Solving an equation and an inequality are similar in the sequencing of steps taken to solve for the variable. The same if-then moves are used to solve for the variable.

They are different because in an equation, you get one solution, but in an inequality, there are an infinite number of solutions.

Discuss the steps to solving the inequality algebraically.

First, collect like terms on each side of the inequality. To isolate the variable, subtract 4 from both sides. Subtracting a value, 4, from each side of the inequality does not change the solution of the

inequality. Continue to isolate the variable by multiplying both sides by 14

. Multiplying a positive value, 14

, to both sides of the inequality does not change the solution of the inequality.


1. Connor went to the county fair with $𝟐𝟐𝟐𝟐.𝟐𝟐𝟒𝟒 in his pocket. He bought a hot dog and drink for $𝟑𝟑.𝟕𝟕𝟐𝟐 and then wanted to spend the rest of his money on ride tickets, which cost $𝟏𝟏.𝟐𝟐𝟐𝟐 each.

a. Write an inequality to represent the total spent where 𝒓𝒓 is the number of tickets purchased.

𝟏𝟏.𝟐𝟐𝟐𝟐𝒓𝒓+ 𝟑𝟑.𝟕𝟕𝟐𝟐 ≤ 𝟐𝟐𝟐𝟐.𝟐𝟐𝟒𝟒

b. Connor wants to use this inequality to determine whether he can purchase 𝟏𝟏𝟒𝟒 tickets. Use substitution to show whether he will have enough money.

𝟏𝟏.𝟐𝟐𝟐𝟐𝒓𝒓+ 𝟑𝟑.𝟕𝟕𝟐𝟐 ≤ 𝟐𝟐𝟐𝟐.𝟐𝟐𝟒𝟒 𝟏𝟏.𝟐𝟐𝟐𝟐(𝟏𝟏𝟒𝟒) + 𝟑𝟑.𝟕𝟕𝟐𝟐 ≤ 𝟐𝟐𝟐𝟐.𝟐𝟐𝟒𝟒

𝟏𝟏𝟐𝟐.𝟐𝟐 + 𝟑𝟑.𝟕𝟕𝟐𝟐 ≤ 𝟐𝟐𝟐𝟐.𝟐𝟐𝟒𝟒 𝟏𝟏𝟐𝟐.𝟐𝟐𝟐𝟐 ≤ 𝟐𝟐𝟐𝟐.𝟐𝟐𝟒𝟒

True

He will have enough money since a purchase of 𝟏𝟏𝟒𝟒 tickets brings his total spending to $𝟏𝟏𝟐𝟐.𝟐𝟐𝟐𝟐.

MP.2

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7•3 Lesson 13

c. What is the total maximum number of tickets he can buy based upon the given information?

𝟏𝟏.𝟐𝟐𝟐𝟐𝒓𝒓+ 𝟑𝟑.𝟕𝟕𝟐𝟐 ≤ 𝟐𝟐𝟐𝟐.𝟐𝟐𝟒𝟒 𝟏𝟏.𝟐𝟐𝟐𝟐𝒓𝒓+ 𝟑𝟑.𝟕𝟕𝟐𝟐 − 𝟑𝟑.𝟕𝟕𝟐𝟐 ≤ 𝟐𝟐𝟐𝟐.𝟐𝟐𝟒𝟒 − 𝟑𝟑.𝟕𝟕𝟐𝟐

𝟏𝟏.𝟐𝟐𝟐𝟐𝒓𝒓+ 𝟒𝟒 ≤ 𝟏𝟏𝟑𝟑.𝟕𝟕𝟐𝟐

�𝟏𝟏

𝟏𝟏.𝟐𝟐𝟐𝟐� (𝟏𝟏.𝟐𝟐𝟐𝟐𝒓𝒓) ≤ �

𝟏𝟏𝟏𝟏.𝟐𝟐𝟐𝟐

� (𝟏𝟏𝟑𝟑.𝟕𝟕𝟐𝟐)

𝒓𝒓 ≤ 𝟏𝟏𝟐𝟐

The maximum number of tickets he can buy is 𝟏𝟏𝟐𝟐.


2. Write and solve an inequality statement to represent the following problem:

On a particular airline, checked bags can weigh no more than 𝟐𝟐𝟒𝟒 pounds. Sally packed 𝟑𝟑𝟐𝟐 pounds of clothes and five identical gifts in a suitcase that weighs 𝟑𝟑 pounds. Write an inequality to represent this situation.

𝒙𝒙: weight of one gift

𝟐𝟐𝒙𝒙 + 𝟑𝟑 + 𝟑𝟑𝟐𝟐 ≤ 𝟐𝟐𝟒𝟒 𝟐𝟐𝒙𝒙 + 𝟒𝟒𝟒𝟒 ≤ 𝟐𝟐𝟒𝟒

𝟐𝟐𝒙𝒙 + 𝟒𝟒𝟒𝟒 − 𝟒𝟒𝟒𝟒 ≤ 𝟐𝟐𝟒𝟒 − 𝟒𝟒𝟒𝟒 𝟐𝟐𝒙𝒙 ≤ 𝟏𝟏𝟒𝟒

�𝟏𝟏𝟐𝟐� (𝟐𝟐𝒙𝒙) ≤ �

𝟏𝟏𝟐𝟐� (𝟏𝟏𝟒𝟒)

𝒙𝒙 ≤ 𝟐𝟐

Each of the 𝟐𝟐 gifts can weigh 𝟐𝟐 pounds or less.

Closing (3 minutes)

How do you know when you need to use an inequality instead of an equation to model a given situation? Is it possible for an inequality to have exactly one solution? Exactly two solutions? Why or why not?


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7•3 Lesson 13

Name ___________________________________________________ Date____________________


Exit Ticket Shaggy earned $7.55 per hour plus an additional $100 in tips waiting tables on Saturday. He earned at least $160 in all. Write an inequality and find the minimum number of hours, to the nearest hour, that Shaggy worked on Saturday.

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7•3 Lesson 13


Shaggy earned $𝟕𝟕.𝟐𝟐𝟐𝟐 per hour plus an additional $𝟏𝟏𝟒𝟒𝟒𝟒 in tips waiting tables on Saturday. He earned at least $𝟏𝟏𝟐𝟐𝟒𝟒 in all. Write an inequality and find the minimum number of hours, to the nearest hour, that Shaggy worked on Saturday.

Let 𝒉𝒉 represent the number of hours worked.

𝟕𝟕.𝟐𝟐𝟐𝟐𝒉𝒉+ 𝟏𝟏𝟒𝟒𝟒𝟒 ≥ 𝟏𝟏𝟐𝟐𝟒𝟒 𝟕𝟕.𝟐𝟐𝟐𝟐𝒉𝒉+ 𝟏𝟏𝟒𝟒𝟒𝟒 − 𝟏𝟏𝟒𝟒𝟒𝟒 ≥ 𝟏𝟏𝟐𝟐𝟒𝟒 − 𝟏𝟏𝟒𝟒𝟒𝟒

𝟕𝟕.𝟐𝟐𝟐𝟐𝒉𝒉 ≥ 𝟐𝟐𝟒𝟒

�𝟏𝟏

𝟕𝟕.𝟐𝟐𝟐𝟐� (𝟕𝟕.𝟐𝟐𝟐𝟐𝒉𝒉) ≥ �

𝟏𝟏𝟕𝟕.𝟐𝟐𝟐𝟐

� (𝟐𝟐𝟒𝟒)

𝒉𝒉 ≥ 𝟕𝟕.𝟒𝟒

If Shaggy earned at least $𝟏𝟏𝟐𝟐𝟒𝟒, he would have worked at least 𝟑𝟑 hours.

Note: The solution shown above is rounded to the nearest tenth. The overall solution, though, is rounded to the nearest hour since that is what the question asks for.


1. Match each problem to the inequality that models it. One choice will be used twice.

c The sum of three times a number and −𝟒𝟒 is greater than 𝟏𝟏𝟕𝟕. a. 𝟑𝟑𝒙𝒙 + −𝟒𝟒 ≥ 𝟏𝟏𝟕𝟕

b The sum of three times a number and −𝟒𝟒 is less than 𝟏𝟏𝟕𝟕. b. 𝟑𝟑𝒙𝒙 + −𝟒𝟒 < 𝟏𝟏𝟕𝟕

d The sum of three times a number and −𝟒𝟒 is at most 𝟏𝟏𝟕𝟕. c. 𝟑𝟑𝒙𝒙 + −𝟒𝟒 > 𝟏𝟏𝟕𝟕

d The sum of three times a number and −𝟒𝟒 is no more than 𝟏𝟏𝟕𝟕. d. 𝟑𝟑𝒙𝒙 + −𝟒𝟒 ≤ 𝟏𝟏𝟕𝟕

a The sum of three times a number and −𝟒𝟒 is at least 𝟏𝟏𝟕𝟕.

2. If 𝒙𝒙 represents a positive integer, find the solutions to the following inequalities.

a. 𝒙𝒙 < 𝟕𝟕

𝒙𝒙 < 𝟕𝟕 or 𝟏𝟏,𝟐𝟐,𝟑𝟑,𝟒𝟒,𝟐𝟐,𝟐𝟐

b. 𝒙𝒙 − 𝟏𝟏𝟐𝟐 < 𝟐𝟐𝟒𝟒

𝒙𝒙 < 𝟑𝟑𝟐𝟐

c. 𝒙𝒙 + 𝟑𝟑 ≤ 𝟏𝟏𝟐𝟐

𝒙𝒙 ≤ 𝟏𝟏𝟐𝟐

d. −𝒙𝒙 > 𝟐𝟐

There are no positive integer solutions.

e. 𝟏𝟏𝟒𝟒 − 𝒙𝒙 > 𝟐𝟐

𝒙𝒙 < 𝟑𝟑

f. −𝒙𝒙 ≥ 𝟐𝟐


g. 𝒙𝒙𝟑𝟑

< 𝟐𝟐

𝒙𝒙 < 𝟐𝟐

h. −𝒙𝒙𝟑𝟑 > 𝟐𝟐


i. 𝟑𝟑 − 𝒙𝒙𝟒𝟒 > 𝟐𝟐

𝒙𝒙 < 𝟒𝟒

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7•3 Lesson 13

3. Recall that the symbol ≠ means not equal to. If 𝒙𝒙 represents a positive integer, state whether each of the following statements is always true, sometimes true, or false.

a. 𝒙𝒙 > 𝟒𝟒

Always true

b. 𝒙𝒙 < 𝟒𝟒

False

c. 𝒙𝒙 > −𝟐𝟐

Always true

d. 𝒙𝒙 > 𝟏𝟏

Sometimes true

e. 𝒙𝒙 ≥ 𝟏𝟏

Always true

f. 𝒙𝒙 ≠ 𝟒𝟒

Always true

g. 𝒙𝒙 ≠ −𝟏𝟏

Always true

h. 𝒙𝒙 ≠ 𝟐𝟐

Sometimes true

4. Twice the smaller of two consecutive integers increased by the larger integer is at least 𝟐𝟐𝟐𝟐.

Model the problem with an inequality, and determine which of the given values 𝟕𝟕, 𝟑𝟑, and/or 𝟒𝟒 are solutions. Then, find the smallest number that will make the inequality true.

𝟐𝟐𝒙𝒙 + 𝒙𝒙 + 𝟏𝟏 ≥ 𝟐𝟐𝟐𝟐

The smallest integer would be 𝟑𝟑.

5.

a. The length of a rectangular fenced enclosure is 𝟏𝟏𝟐𝟐 feet more than the width. If Farmer Dan has 𝟏𝟏𝟒𝟒𝟒𝟒 feet of fencing, write an inequality to find the dimensions of the rectangle with the largest perimeter that can be created using 𝟏𝟏𝟒𝟒𝟒𝟒 feet of fencing.

Let 𝟑𝟑 represent the width of the fenced enclosure.

𝟑𝟑 + 𝟏𝟏𝟐𝟐: length of the fenced enclosure

𝟑𝟑 +𝟑𝟑 +𝟑𝟑 + 𝟏𝟏𝟐𝟐 +𝟑𝟑 + 𝟏𝟏𝟐𝟐 ≤ 𝟏𝟏𝟒𝟒𝟒𝟒 𝟒𝟒𝟑𝟑 + 𝟐𝟐𝟒𝟒 ≤ 𝟏𝟏𝟒𝟒𝟒𝟒

For 𝒙𝒙 = 𝟕𝟕:

𝟐𝟐𝒙𝒙 + 𝒙𝒙 + 𝟏𝟏 ≥ 𝟐𝟐𝟐𝟐

𝟐𝟐(𝟕𝟕) + 𝟕𝟕 + 𝟏𝟏 ≥ 𝟐𝟐𝟐𝟐 𝟏𝟏𝟒𝟒 + 𝟕𝟕 + 𝟏𝟏 ≥ 𝟐𝟐𝟐𝟐

𝟐𝟐𝟐𝟐 ≥ 𝟐𝟐𝟐𝟐 False

For 𝒙𝒙 = 𝟑𝟑:

𝟐𝟐𝒙𝒙 + 𝒙𝒙 + 𝟏𝟏 ≥ 𝟐𝟐𝟐𝟐

𝟐𝟐(𝟑𝟑) + 𝟑𝟑 + 𝟏𝟏 ≥ 𝟐𝟐𝟐𝟐 𝟏𝟏𝟐𝟐 + 𝟑𝟑 + 𝟏𝟏 ≥ 𝟐𝟐𝟐𝟐

𝟐𝟐𝟐𝟐 ≥ 𝟐𝟐𝟐𝟐 True

For 𝒙𝒙 = 𝟒𝟒:

𝟐𝟐𝒙𝒙 + 𝒙𝒙 + 𝟏𝟏 ≥ 𝟐𝟐𝟐𝟐

𝟐𝟐(𝟒𝟒) + 𝟒𝟒 + 𝟏𝟏 ≥ 𝟐𝟐𝟐𝟐 𝟏𝟏𝟑𝟑 + 𝟒𝟒 + 𝟏𝟏 ≥ 𝟐𝟐𝟐𝟐

𝟐𝟐𝟑𝟑 ≥ 𝟐𝟐𝟐𝟐 True

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7•3 Lesson 13

b. What are the dimensions of the rectangle with the largest perimeter? What is the area enclosed by this rectangle?

𝟒𝟒𝟑𝟑 + 𝟐𝟐𝟒𝟒 ≤ 𝟏𝟏𝟒𝟒𝟒𝟒 𝟒𝟒𝟑𝟑 + 𝟐𝟐𝟒𝟒 − 𝟐𝟐𝟒𝟒 ≤ 𝟏𝟏𝟒𝟒𝟒𝟒 − 𝟐𝟐𝟒𝟒

𝟒𝟒𝟑𝟑 + 𝟒𝟒 ≤ 𝟕𝟕𝟐𝟐

�𝟏𝟏𝟒𝟒� (𝟒𝟒𝟑𝟑) ≤ �

𝟏𝟏𝟒𝟒� (𝟕𝟕𝟐𝟐)

𝟑𝟑 ≤ 𝟏𝟏𝟒𝟒

Maximum width is 𝟏𝟏𝟒𝟒 feet.

Maximum length is 𝟑𝟑𝟏𝟏 feet.

Maximum area: 𝑨𝑨 = 𝒍𝒍𝟑𝟑

𝑨𝑨 = (𝟏𝟏𝟒𝟒)(𝟑𝟑𝟏𝟏) 𝑨𝑨 = 𝟐𝟐𝟑𝟑𝟒𝟒

The area is 𝟐𝟐𝟑𝟑𝟒𝟒 𝐟𝐟𝐭𝐭𝟐𝟐.

6. At most, Kyle can spend $𝟐𝟐𝟒𝟒 on sandwiches and chips for a picnic. He already bought chips for $𝟐𝟐 and will buy sandwiches that cost $𝟒𝟒.𝟐𝟐𝟒𝟒 each. Write and solve an inequality to show how many sandwiches he can buy. Show your work, and interpret your solution.

Let 𝒔𝒔 represent the number of sandwiches.

𝟒𝟒.𝟐𝟐𝟒𝟒𝒔𝒔 + 𝟐𝟐 ≤ 𝟐𝟐𝟒𝟒

𝟒𝟒.𝟐𝟐𝟒𝟒𝒔𝒔 + 𝟐𝟐 − 𝟐𝟐 ≤ 𝟐𝟐𝟒𝟒 − 𝟐𝟐 𝟒𝟒.𝟐𝟐𝟒𝟒𝒔𝒔 ≤ 𝟒𝟒𝟒𝟒

�𝟏𝟏

𝟒𝟒.𝟐𝟐𝟒𝟒� (𝟒𝟒.𝟐𝟐𝟒𝟒𝒔𝒔) ≤ �

𝟏𝟏𝟒𝟒.𝟐𝟐𝟒𝟒

� (𝟒𝟒𝟒𝟒)

𝒔𝒔 ≤ 𝟒𝟒𝟕𝟕𝟒𝟒

At most, Kyle can buy 𝟒𝟒 sandwiches with $𝟐𝟐𝟒𝟒.

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Lesson 14: Solving Inequalities

7•3 Lesson 14


Student Outcomes

Students solve word problems leading to inequalities that compare 𝑝𝑝𝑝𝑝 + 𝑞𝑞 and 𝑟𝑟, where 𝑝𝑝, 𝑞𝑞, and 𝑟𝑟 are specific rational numbers.

Students interpret the solutions in the context of the problem.

Classwork

Opening (1 minute)

Start the lesson by discussing some summertime events that students may attend. One event may be a carnival or a fair.

The problems that students complete today are all about a local carnival in their town that lasts 5 12 days.


The annual County Carnival is being held this summer and will last 𝟓𝟓𝟏𝟏𝟐𝟐 days. Use this information and the other given information to answer each problem.

You are the owner of the biggest and newest roller coaster called the Gentle Giant. The roller coaster costs $𝟔𝟔 to ride. The operator of the ride must pay $𝟐𝟐𝟐𝟐𝟐𝟐 per day for the ride rental and $𝟔𝟔𝟓𝟓 per day for a safety inspection. If you want to make a profit of at least $𝟏𝟏,𝟐𝟐𝟐𝟐𝟐𝟐 each day, what is the minimum number of people that must ride the roller coaster?

Write an inequality that can be used to find the minimum number of people, 𝒑𝒑, which must ride the roller coaster each day to make the daily profit.

𝟔𝟔𝒑𝒑 − 𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟔𝟔𝟓𝟓 ≥ 𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐

Solve the inequality.

𝟔𝟔𝒑𝒑 − 𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟔𝟔𝟓𝟓 ≥ 𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐

𝟔𝟔𝒑𝒑 − 𝟐𝟐𝟔𝟔𝟓𝟓 ≥ 𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐 𝟔𝟔𝒑𝒑 − 𝟐𝟐𝟔𝟔𝟓𝟓+ 𝟐𝟐𝟔𝟔𝟓𝟓 ≥ 𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟔𝟔𝟓𝟓

𝟔𝟔𝒑𝒑 + 𝟐𝟐 ≥ 𝟏𝟏𝟐𝟐𝟔𝟔𝟓𝟓

�𝟏𝟏𝟔𝟔� (𝟔𝟔𝒑𝒑) ≥ �

𝟏𝟏𝟔𝟔� (𝟏𝟏𝟐𝟐𝟔𝟔𝟓𝟓)

𝒑𝒑 ≥ 𝟐𝟐𝟏𝟏𝟐𝟐𝟓𝟓𝟔𝟔

Scaffolding:

Use integers to lead to the inequality.

What would be the profit if 10 people rode the roller coaster? 6(10) − 200 − 65 = −205

What does the answer mean within the context of the problem?

There was not a profit; the owner lost $205.

What would be the profit if 50 people rode the roller coaster? 6(50) − 200 − 65 = 35


There owner earned $35.

What would be the profit if 200 people rode the roller coaster? 6(200) − 200 − 65 = 935


The owner earned $935.

Recall that profit is the revenue (money received) less the expenses (money spent).

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7•3 Lesson 14

Interpret the solution.

There needs to be a minimum of 𝟐𝟐𝟏𝟏𝟏𝟏 people to ride the roller coaster every day to make a daily profit of at least $𝟏𝟏,𝟐𝟐𝟐𝟐𝟐𝟐.

Discussion

Recall the formula for profit as revenue − expenses. In this example, what expression represents the revenue, and what expression represents the expenses?

The revenue is the money coming in. This would be $6 per person.

The expenses are the money spent or going out. This would be the daily cost of renting the ride, $200, and the daily cost of safety inspections, $65.

Why was the inequality ≥ used?

The owner would be satisfied if the profit was at least $1,000 or more. The phrase at least means greater than or equal to.

Was it necessary to flip or reverse the inequality sign? Explain why or why not.

No, it was not necessary to reverse the inequality sign. This is because we did not multiply or divide by a negative number.

Describe the if-then moves used in solving the inequality.

After combining like terms, 265 was added to both sides. Adding a number to both sides of the

inequality does not change the solution of the inequality. Lastly 16 was multiplied to both sides to isolate the variable.

Why is the answer 211 people versus 210 people?

The answer has to be greater than or equal to 210 56 people. You cannot have 56 of a person. If only

210 people purchased tickets, the profit would be $995, which is less than $1,000. Therefore, we round up to assure the profit of at least $1,000.

The variable 𝑝𝑝 represents the number of people who ride the roller coaster each day. Explain the importance of clearly defining 𝑝𝑝 as people riding the roller coaster per day versus people who ride it the entire carnival time. How would the inequality change if 𝑝𝑝 were the number of people who rode the roller coaster the entire time?

Since the expenses and profit were given as daily figures, then 𝑝𝑝 would represent the number of people who rode the ride daily. The units have to be the same. If 𝑝𝑝 were for the entire time the carnival was in

town, then the desired profit would be $1,000 for the entire 5 12 days instead of daily. However, the

expenses were given as daily costs. Therefore, to determine the number of tickets that need to be sold to achieve a profit of at least $1,000 for the entire time the carnival is in town, we will need to calculate

the total expense by multiplying the daily expenses by 5 12. The new inequality would be

6𝑝𝑝 − 5.5(265) ≥ 1,000, which would change the answer to 410 people overall.

What if the expenses were charged for a whole day versus a half day? How would that change the inequality and answer?

The expenses would be multiplied by 6, which would change the answer to 432 people. What if the intended profit was still $1,000 per day, but 𝑝𝑝 was the number of people who rode the roller

coaster the entire time the carnival was in town?

The expenses and desired profit would be multiplied by 5.5. The answer would change to a total of 1,160 people.

MP.2 &

MP.6

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Example 1

A youth summer camp has budgeted $𝟐𝟐,𝟐𝟐𝟐𝟐𝟐𝟐 for the campers to attend the carnival. The cost for each camper is $𝟏𝟏𝟏𝟏.𝟗𝟗𝟓𝟓, which includes general admission to the carnival and two meals. The youth summer camp must also pay $𝟐𝟐𝟓𝟓𝟐𝟐 for the chaperones to attend the carnival and $𝟑𝟑𝟓𝟓𝟐𝟐 for transportation to and from the carnival. What is the greatest number of campers who can attend the carnival if the camp must stay within its budgeted amount?

Let 𝒄𝒄 represent the number of campers to attend the carnival.

𝟏𝟏𝟏𝟏.𝟗𝟗𝟓𝟓𝒄𝒄 + 𝟐𝟐𝟓𝟓𝟐𝟐+ 𝟑𝟑𝟓𝟓𝟐𝟐 ≤ 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏.𝟗𝟗𝟓𝟓𝒄𝒄 + 𝟔𝟔𝟐𝟐𝟐𝟐 ≤ 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏.𝟗𝟗𝟓𝟓𝒄𝒄 + 𝟔𝟔𝟐𝟐𝟐𝟐 − 𝟔𝟔𝟐𝟐𝟐𝟐 ≤ 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟔𝟔𝟐𝟐𝟐𝟐

𝟏𝟏𝟏𝟏.𝟗𝟗𝟓𝟓𝒄𝒄 ≤ 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐

�𝟏𝟏

𝟏𝟏𝟏𝟏.𝟗𝟗𝟓𝟓� (𝟏𝟏𝟏𝟏.𝟗𝟗𝟓𝟓𝒄𝒄) ≤ �

𝟏𝟏𝟏𝟏𝟏𝟏.𝟗𝟗𝟓𝟓

� (𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐)

𝒄𝒄 ≤ 𝟏𝟏𝟏𝟏.𝟗𝟗𝟗𝟗

In order for the camp to stay in budget, the greatest number of campers who can attend the carnival is 𝟏𝟏𝟏𝟏 campers.

Why is the inequality ≤ used?

The camp can spend less than the budgeted amount or the entire amount, but cannot spend more.

Describe the if-then moves used to solve the inequality.

Once like terms were collected, then the goal was to isolate the variable to get 0s and 1s. If a number, such as 600 is subtracted from each side of an inequality, then the solution of the inequality does not

change. If a positive number, 117.95, is multiplied to each side of the inequality, then the solution of the

inequality does not change.

Why did we round down instead of rounding up? In the context of the problem, the number of campers has to be less than 77.99 campers. Rounding up

to 78 would be greater than 77.99; thus, we rounded down.

How can the equation be written to clear the decimals, resulting in an inequality with integer coefficients? Write the equivalent inequality.

Since the decimal terminates in the 100th place, to clear the decimals we can multiply every term by 100. The equivalent inequality would be 1,795 + 60,000 ≤ 200,000.

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Example 2

The carnival owner pays the owner of an exotic animal exhibit $𝟔𝟔𝟓𝟓𝟐𝟐 for the entire time the exhibit is displayed. The owner of the exhibit has no other expenses except for a daily insurance cost. If the owner of the animal exhibit wants to

make more than $𝟓𝟓𝟐𝟐𝟐𝟐 in profits for the 𝟓𝟓𝟏𝟏𝟐𝟐 days, what is the greatest daily insurance rate he can afford to pay?

Let 𝒊𝒊 represent the daily insurance cost.

𝟔𝟔𝟓𝟓𝟐𝟐 − 𝟓𝟓.𝟓𝟓𝒊𝒊 > 𝟓𝟓𝟐𝟐𝟐𝟐 −𝟓𝟓.𝟓𝟓𝒊𝒊 + 𝟔𝟔𝟓𝟓𝟐𝟐 − 𝟔𝟔𝟓𝟓𝟐𝟐 > 𝟓𝟓𝟐𝟐𝟐𝟐 − 𝟔𝟔𝟓𝟓𝟐𝟐

−𝟓𝟓.𝟓𝟓𝒊𝒊 + 𝟐𝟐 > −𝟏𝟏𝟓𝟓𝟐𝟐

�𝟏𝟏

−𝟓𝟓.𝟓𝟓� (−𝟓𝟓.𝟓𝟓𝒊𝒊) > �

𝟏𝟏−𝟓𝟓.𝟓𝟓

� (−𝟏𝟏𝟓𝟓𝟐𝟐)

𝒊𝒊 < 𝟐𝟐𝟏𝟏.𝟐𝟐𝟏𝟏

The maximum daily cost the owner can pay for insurance is $𝟐𝟐𝟏𝟏.𝟐𝟐𝟏𝟏.

Encourage students to verbalize the if-then moves used to obtain a solution.

Since the desired profit was greater than (>) $500, the inequality used was >. Why, then, is the answer 𝑖𝑖 < 27.27?

When solving the inequality, we multiplied both sides by a negative number. When you multiply or divide by a negative number, the inequality is NOT preserved, and it is reversed.

Why was the answer rounded to 2 decimal places? Since 𝑖𝑖 represents the daily cost, in cents, then when we are working with money, the decimal is

rounded to the hundredth place, or two decimal places.

The answer is 𝑖𝑖 < 27.27. Notice that the inequality is not less than or equal to. The largest number less than 27.27 is 27.26. However, the daily cost is still $27.27. Why is the maximum daily cost $27.27 and not $27.26?

The profit had to be more than $500, not equal to $500. The precise answer is 27.27272727. Since the answer is rounded to $27.27, the actual profit, when 27.27 is substituted into the expression, would be 500.01, which is greater than $500.

Write an equivalent inequality clearing the decimals. 6,500 − 55𝑖𝑖 > 5,000

Why do we multiply by 10 and not 100 to clear the decimals? The smallest decimal terminates in the tenths place.

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Example 3

Several vendors at the carnival sell products and advertise their businesses. Shane works for a recreational company that sells ATVs, dirt bikes, snowmobiles, and motorcycles. His boss paid him $𝟓𝟓𝟐𝟐𝟐𝟐 for working all of the days at the carnival plus 𝟓𝟓% commission on all of the sales made at the carnival. What was the minimum amount of sales Shane needed to make if he earned more than $𝟏𝟏,𝟓𝟓𝟐𝟐𝟐𝟐?

Let 𝒔𝒔 represent the sales, in dollars, made during the carnival.

𝟓𝟓𝟐𝟐𝟐𝟐 +𝟓𝟓𝟏𝟏𝟐𝟐𝟐𝟐

𝒔𝒔 > 𝟏𝟏,𝟓𝟓𝟐𝟐𝟐𝟐

𝟓𝟓𝟏𝟏𝟐𝟐𝟐𝟐

𝒔𝒔 + 𝟓𝟓𝟐𝟐𝟐𝟐 > 𝟏𝟏,𝟓𝟓𝟐𝟐𝟐𝟐

𝟓𝟓𝟏𝟏𝟐𝟐𝟐𝟐

𝒔𝒔 + 𝟓𝟓𝟐𝟐𝟐𝟐 − 𝟓𝟓𝟐𝟐𝟐𝟐 > 𝟏𝟏,𝟓𝟓𝟐𝟐𝟐𝟐 − 𝟓𝟓𝟐𝟐𝟐𝟐

𝟓𝟓𝟏𝟏𝟐𝟐𝟐𝟐

𝒔𝒔 + 𝟐𝟐 > 𝟏𝟏,𝟐𝟐𝟐𝟐𝟐𝟐

�𝟏𝟏𝟐𝟐𝟐𝟐𝟓𝟓��

𝟓𝟓𝟏𝟏𝟐𝟐𝟐𝟐

𝒔𝒔� > �𝟏𝟏𝟐𝟐𝟐𝟐𝟓𝟓� (𝟏𝟏,𝟐𝟐𝟐𝟐𝟐𝟐)

𝒔𝒔 > 𝟐𝟐𝟐𝟐,𝟐𝟐𝟐𝟐𝟐𝟐

The sales had to be more than $𝟐𝟐𝟐𝟐,𝟐𝟐𝟐𝟐𝟐𝟐 for Shane to earn more than $𝟏𝟏,𝟓𝟓𝟐𝟐𝟐𝟐.

Encourage students to verbalize the if-then moves used in obtaining the solution.

Recall from Module 2 how to work with a percent. Percents are out of 100, so what fraction or decimal represents 5%?

5100

or 0.05

How can we write an equivalent inequality containing only integer coefficients and constant terms? Write the equivalent inequality.

Every term can be multiplied by the common denominator of the fraction. In this case, the only denominator is 100. After clearing the fraction, the equivalent inequality is 50,000 + 5𝑠𝑠 > 150,000.

Solve the new inequality. 50,000 + 5𝑠𝑠 > 150,000

5𝑠𝑠 + 50,000 − 50,000 > 150,000 − 50,000 5𝑠𝑠 + 0 > 100,000

�15� (5𝑠𝑠) > �

15� (100,000)

𝑠𝑠 > 20,000

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Closing (3 minutes)

What did all of the situations that required an inequality to solve have in common?

How is a solution of an inequality interpreted?


Lesson Summary

The key to solving inequalities is to use if-then moves to make 𝟐𝟐’s and 𝟏𝟏’s to get the inequality into the form 𝒙𝒙 > 𝒄𝒄 or 𝒙𝒙 < 𝒄𝒄 where 𝒄𝒄 is a number. Adding or subtracting opposites will make 𝟐𝟐’s. According to the if-then move, any number that is added to or subtracted from each side of an inequality does not change the solution to the inequality. Multiplying and dividing numbers makes 𝟏𝟏’s. When each side of an inequality is multiplied by or divided by a positive number, the sign of the inequality is not reversed. However, when each side of an inequality is multiplied by or divided by a negative number, the sign of the inequality is reversed.

Given inequalities containing decimals, equivalent inequalities can be created which have only integer coefficients and constant terms by repeatedly multiplying every term by ten until all coefficients and constant terms are integers.

Given inequalities containing fractions, equivalent inequalities can be created which have only integer coefficients and constant terms by multiplying every term by the least common multiple of the values in the denominators.

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Name ___________________________________________________ Date____________________


Exit Ticket Games at the carnival cost $3 each. The prizes awarded to winners cost $145.65. How many games must be played to make at least $50?

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Games at the carnival cost $𝟑𝟑 each. The prizes awarded to winners cost $𝟏𝟏𝟏𝟏𝟓𝟓.𝟔𝟔𝟓𝟓. How many games must be played to make at least $𝟓𝟓𝟐𝟐?

Let 𝒈𝒈 represent the number of games played.

𝟑𝟑𝒈𝒈 − 𝟏𝟏𝟏𝟏𝟓𝟓.𝟔𝟔𝟓𝟓 ≥ 𝟓𝟓𝟐𝟐 𝟑𝟑𝒈𝒈 − 𝟏𝟏𝟏𝟏𝟓𝟓.𝟔𝟔𝟓𝟓+ 𝟏𝟏𝟏𝟏𝟓𝟓.𝟔𝟔𝟓𝟓 ≥ 𝟓𝟓𝟐𝟐+ 𝟏𝟏𝟏𝟏𝟓𝟓.𝟔𝟔𝟓𝟓

𝟑𝟑𝒈𝒈+ 𝟐𝟐 ≥ 𝟏𝟏𝟗𝟗𝟓𝟓.𝟔𝟔𝟓𝟓

�𝟏𝟏𝟑𝟑� (𝟑𝟑𝒈𝒈) ≥ �

𝟏𝟏𝟑𝟑� (𝟏𝟏𝟗𝟗𝟓𝟓.𝟔𝟔𝟓𝟓)

𝒈𝒈 ≥ 𝟔𝟔𝟓𝟓.𝟐𝟐𝟏𝟏𝟏𝟏

There must be at least 𝟔𝟔𝟔𝟔 games played to make at least $𝟓𝟓𝟐𝟐.


1. As a salesperson, Jonathan is paid $𝟓𝟓𝟐𝟐 per week plus 𝟑𝟑% of the total amount he sells. This week, he wants to earn at least $𝟏𝟏𝟐𝟐𝟐𝟐. Write an inequality with integer coefficients for the total sales needed to earn at least $𝟏𝟏𝟐𝟐𝟐𝟐, and describe what the solution represents.

Let the variable 𝒑𝒑 represent the purchase amount.

𝟓𝟓𝟐𝟐 +𝟑𝟑𝟏𝟏𝟐𝟐𝟐𝟐

𝒑𝒑 ≥ 𝟏𝟏𝟐𝟐𝟐𝟐

𝟑𝟑𝟏𝟏𝟐𝟐𝟐𝟐

𝒑𝒑 + 𝟓𝟓𝟐𝟐 ≥ 𝟏𝟏𝟐𝟐𝟐𝟐

(𝟏𝟏𝟐𝟐𝟐𝟐) �𝟑𝟑𝟏𝟏𝟐𝟐𝟐𝟐

𝒑𝒑�+ 𝟏𝟏𝟐𝟐𝟐𝟐(𝟓𝟓𝟐𝟐) ≥ 𝟏𝟏𝟐𝟐𝟐𝟐(𝟏𝟏𝟐𝟐𝟐𝟐)

𝟑𝟑𝒑𝒑 + 𝟓𝟓𝟐𝟐𝟐𝟐𝟐𝟐 ≥ 𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐

𝟑𝟑𝒑𝒑 + 𝟓𝟓𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟓𝟓𝟐𝟐𝟐𝟐𝟐𝟐 ≥ 𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟓𝟓𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝒑𝒑 + 𝟐𝟐 ≥ 𝟓𝟓𝟐𝟐𝟐𝟐𝟐𝟐

�𝟏𝟏𝟑𝟑� (𝟑𝟑𝒑𝒑) ≥ �

𝟏𝟏𝟑𝟑� (𝟓𝟓𝟐𝟐𝟐𝟐𝟐𝟐)

𝒑𝒑 ≥ 𝟏𝟏𝟔𝟔𝟔𝟔𝟔𝟔𝟐𝟐𝟑𝟑

Jonathan must sell $𝟏𝟏,𝟔𝟔𝟔𝟔𝟔𝟔.𝟔𝟔𝟏𝟏 in total purchases.

2. Systolic blood pressure is the higher number in a blood pressure reading. It is measured as the heart muscle contracts. Heather was with her grandfather when he had his blood pressure checked. The nurse told him that the upper limit of his systolic blood pressure is equal to half his age increased by 𝟏𝟏𝟏𝟏𝟐𝟐.

a. 𝒂𝒂 is the age in years, and 𝒑𝒑 is the systolic blood pressure in millimeters of mercury (mmHg). Write an inequality to represent this situation.

𝒑𝒑 ≤𝟏𝟏𝟐𝟐𝒂𝒂 + 𝟏𝟏𝟏𝟏𝟐𝟐

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b. Heather’s grandfather is 𝟏𝟏𝟔𝟔 years old. What is normal for his systolic blood pressure?

𝒑𝒑 ≤ 𝟏𝟏𝟐𝟐𝒂𝒂 + 𝟏𝟏𝟏𝟏𝟐𝟐, where 𝒂𝒂 = 𝟏𝟏𝟔𝟔.

𝒑𝒑 ≤𝟏𝟏𝟐𝟐

(𝟏𝟏𝟔𝟔) + 𝟏𝟏𝟏𝟏𝟐𝟐

𝒑𝒑 ≤ 𝟑𝟑𝟑𝟑+ 𝟏𝟏𝟏𝟏𝟐𝟐

𝒑𝒑 ≤ 𝟏𝟏𝟏𝟏𝟑𝟑

A systolic blood pressure for his age is normal if it is at most 𝟏𝟏𝟏𝟏𝟑𝟑 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎.

3. Traci collects donations for a dance marathon. One group of sponsors will donate a total of $𝟔𝟔 for each hour she dances. Another group of sponsors will donate $𝟏𝟏𝟓𝟓 no matter how long she dances. What number of hours, to the nearest minute, should Traci dance if she wants to raise at least $𝟏𝟏,𝟐𝟐𝟐𝟐𝟐𝟐?

Let the variable 𝒉𝒉 represent the number of hours Traci dances.

𝟔𝟔𝒉𝒉 + 𝟏𝟏𝟓𝟓 ≥ 𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐 𝟔𝟔𝒉𝒉 + 𝟏𝟏𝟓𝟓 − 𝟏𝟏𝟓𝟓 ≥ 𝟏𝟏𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟓𝟓

𝟔𝟔𝒉𝒉 + 𝟐𝟐 ≥ 𝟗𝟗𝟐𝟐𝟓𝟓

�𝟏𝟏𝟔𝟔� (𝟔𝟔𝒉𝒉) ≥ �

𝟏𝟏𝟔𝟔� (𝟗𝟗𝟐𝟐𝟓𝟓)

𝒉𝒉 ≥ 𝟏𝟏𝟓𝟓𝟏𝟏𝟏𝟏𝟔𝟔

Traci would have to dance at least 𝟏𝟏𝟓𝟓𝟏𝟏 hours and 𝟏𝟏𝟐𝟐 minutes.

4. Jack’s age is three years more than twice the age of his younger brother, Jimmy. If the sum of their ages is at most 𝟏𝟏𝟑𝟑, find the greatest age that Jimmy could be.

Let the variable 𝒋𝒋 represent Jimmy’s age in years.

Then, the expression 𝟑𝟑 + 𝟐𝟐𝒋𝒋 represents Jack’s age in years.

𝒋𝒋 + 𝟑𝟑+ 𝟐𝟐𝒋𝒋 ≤ 𝟏𝟏𝟑𝟑

𝟑𝟑𝒋𝒋+ 𝟑𝟑 ≤ 𝟏𝟏𝟑𝟑 𝟑𝟑𝒋𝒋 + 𝟑𝟑 − 𝟑𝟑 ≤ 𝟏𝟏𝟑𝟑 − 𝟑𝟑

𝟑𝟑𝒋𝒋 ≤ 𝟏𝟏𝟓𝟓

�𝟏𝟏𝟑𝟑� (𝟑𝟑𝒋𝒋) ≤ �

𝟏𝟏𝟑𝟑� (𝟏𝟏𝟓𝟓)

𝒋𝒋 ≤ 𝟓𝟓

Jimmy’s age is 𝟓𝟓 years or less.

5. Brenda has $𝟓𝟓𝟐𝟐𝟐𝟐 in her bank account. Every week she withdraws $𝟏𝟏𝟐𝟐 for miscellaneous expenses. How many weeks can she withdraw the money if she wants to maintain a balance of a least $𝟐𝟐𝟐𝟐𝟐𝟐?

Let the variable 𝒘𝒘 represent the number of weeks.

𝟓𝟓𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟐𝟐𝒘𝒘 ≥ 𝟐𝟐𝟐𝟐𝟐𝟐

𝟓𝟓𝟐𝟐𝟐𝟐 − 𝟓𝟓𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟐𝟐𝒘𝒘 ≥ 𝟐𝟐𝟐𝟐𝟐𝟐 − 𝟓𝟓𝟐𝟐𝟐𝟐 −𝟏𝟏𝟐𝟐𝒘𝒘 ≥ −𝟑𝟑𝟐𝟐𝟐𝟐

�−𝟏𝟏𝟏𝟏𝟐𝟐� (−𝟏𝟏𝟐𝟐𝒘𝒘) ≤ �−

𝟏𝟏𝟏𝟏𝟐𝟐� (−𝟑𝟑𝟐𝟐𝟐𝟐)

𝒘𝒘 ≤ 𝟏𝟏.𝟓𝟓

$𝟏𝟏𝟐𝟐 can be withdrawn from the account for seven weeks if she wants to maintain a balance of at least $𝟐𝟐𝟐𝟐𝟐𝟐.

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6. A scooter travels 𝟏𝟏𝟐𝟐 miles per hour faster than an electric bicycle. The scooter traveled for 𝟑𝟑 hours, and the bicycle

traveled for 𝟓𝟓𝟏𝟏𝟐𝟐 hours. Altogether, the scooter and bicycle traveled no more than 𝟐𝟐𝟑𝟑𝟓𝟓 miles. Find the maximum speed of each.

Speed Time Distance

Scooter 𝒙𝒙 + 𝟏𝟏𝟐𝟐 𝟑𝟑 𝟑𝟑(𝒙𝒙 + 𝟏𝟏𝟐𝟐)

Bicycle 𝒙𝒙 𝟓𝟓𝟏𝟏𝟐𝟐

𝟓𝟓𝟏𝟏𝟐𝟐𝒙𝒙

𝟑𝟑(𝒙𝒙 + 𝟏𝟏𝟐𝟐) + 𝟓𝟓𝟏𝟏𝟐𝟐𝒙𝒙 ≤ 𝟐𝟐𝟑𝟑𝟓𝟓

𝟑𝟑𝒙𝒙 + 𝟑𝟑𝟐𝟐 + 𝟓𝟓𝟏𝟏𝟐𝟐𝒙𝒙 ≤ 𝟐𝟐𝟑𝟑𝟓𝟓

𝟑𝟑𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟑𝟑𝟐𝟐 ≤ 𝟐𝟐𝟑𝟑𝟓𝟓

𝟑𝟑𝟏𝟏𝟐𝟐𝒙𝒙 + 𝟑𝟑𝟐𝟐 − 𝟑𝟑𝟐𝟐 ≤ 𝟐𝟐𝟑𝟑𝟓𝟓 − 𝟑𝟑𝟐𝟐

𝟑𝟑𝟏𝟏𝟐𝟐𝒙𝒙 ≤ 𝟐𝟐𝟓𝟓𝟓𝟓

𝟏𝟏𝟏𝟏𝟐𝟐𝒙𝒙 ≤ 𝟐𝟐𝟓𝟓𝟓𝟓

�𝟐𝟐𝟏𝟏𝟏𝟏� �𝟏𝟏𝟏𝟏𝟐𝟐𝒙𝒙� ≤ (𝟐𝟐𝟓𝟓𝟓𝟓) �

𝟐𝟐𝟏𝟏𝟏𝟏�

𝒙𝒙 ≤ 𝟑𝟑𝟐𝟐

The maximum speed the bicycle traveled was 𝟑𝟑𝟐𝟐 miles per hour, and the maximum speed the scooter traveled was 𝟏𝟏𝟐𝟐 miles per hour.

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Lesson 15: Graphing Solutions to Inequalities

7•3 Lesson 15


Student Outcomes

Students graph solutions to inequalities taking care to interpret the solutions in the context of the problem.

Classwork

Rapid White Board Exchange (10 minutes): Inequalities

Students complete a rapid whiteboard exchange where they practice their knowledge of solving linear inequalities in the form 𝑝𝑝𝑝𝑝 + 𝑞𝑞 > 𝑟𝑟 and 𝑝𝑝(𝑝𝑝 + 𝑞𝑞) > 𝑟𝑟.

Discussion/Exercise 1 (10 minutes) Exercise 1

1. Two identical cars need to fit into a small garage. The opening is 𝟐𝟐𝟐𝟐 feet 𝟔𝟔 inches wide, and there must be at least 𝟐𝟐 feet 𝟔𝟔 inches of clearance between the cars and between the edges of the garage. How wide can the cars be?

Encourage students to begin by drawing a diagram to illustrate the problem. A sample diagram is as follows:

Have students try to find all of the widths that the cars could be. Challenge them to name one more width than the person next to them. While they name the widths, plot the widths on a number line at the front of the class to demonstrate the shading. Before plotting the widths, ask if the circle should be open or closed as a quick review of graphing inequalities. Ultimately, the graph should be

MP.7

𝟎𝟎 𝟎𝟎.𝟓𝟓 𝟏𝟏 𝟏𝟏.𝟓𝟓 𝟐𝟐 𝟐𝟐.𝟓𝟓 𝟐𝟐 𝟐𝟐.𝟓𝟓 𝟒𝟒 𝟔𝟔.𝟓𝟓 𝟕𝟕 𝟕𝟕.𝟓𝟓 𝟖𝟖 𝟖𝟖.𝟓𝟓 𝟗𝟗 𝟗𝟗.𝟓𝟓 𝟏𝟏𝟎𝟎 𝟒𝟒.𝟓𝟓 𝟓𝟓 𝟓𝟓.𝟓𝟓 𝟔𝟔

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7•3 Lesson 15

Describe how to find the width of each car.

To find the width of each car, I subtract the minimum amount of space needed on either side of each car and in between the cars from the total length. Altogether, the amount of space needed was

3(3.5 ft. ) or 10.5 ft. Then, I divided the result, 23.5 − 10.5 = 13, by 2 since there were 2 cars. The

answer would be no more than 132 ft. or 6.5 ft.

Did you take an algebraic approach to finding the width of each car or an arithmetic approach? Explain.

Answers will vary.

If arithmetic was used, ask, “If 𝑤𝑤 is the width of one car, write an inequality that can be used to find all possible values of 𝑤𝑤.” 2𝑤𝑤 + 10.5 ≤ 23.5

Why is an inequality used instead of an equation?

Since the minimum amount of space between the cars and each side of the garage is at least 3 feet 6 inches, which equals 3.5 ft., the space could be larger than 3 feet 6 inches. If so, then the width of the cars would be smaller. Since the width in between the cars and on the sides is not exactly 3 feet 6 inches, and it could be more, then there are many possible car widths. An inequality will give all possible car widths.

If an algebraic approach was used initially, ask, “How is the work shown in solving the inequality similar to the arithmetic approach?”

The steps to solving the inequality are the same as in an arithmetic approach. First, determine the total minimum amount of space needed by multiplying 3 by 3.5. Then, subtract 10.5 from the total of 23.5 and divide by 2.

What happens if the width of each car is less than 6.5 feet?

The amount of space between the cars and on either side of the car and garage is more then 3 feet 6 inches.

What happens if the width of each car is exactly 6.5 feet? The amount of space between the cars and on either side of the car and garage is exactly 3 feet

6 inches.

What happens if the width of each car is more than 6.5 feet?

The amount of space between the cars and on either side of the car and garage is less than 3 feet 6 inches.

How many possible car widths are there?

Any infinite number of possible widths.

What assumption is being made?

The assumption made is that the width of the car is greater than 0 feet. The graph illustrates all possible values less than 6.5 feet, but in the context of the problem, we know that the width of the car must be greater than 0 feet.

Since we have determined there is an infinite amount, how can we illustrate this on a number line?

Illustrate by drawing a graph with a closed circle on 6.5 and an arrow drawn to the left.

What if 6.5 feet could not be a width, but all other possible measures less than 6.5 can be a possible width; how would the graph be different?

The graph would have an open circle on 6.5 and an arrow drawn to the left.

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7•3 Lesson 15

𝟓𝟓 𝟔𝟔 𝟕𝟕 𝟖𝟖 𝟏𝟏𝟏𝟏 𝟏𝟏𝟐𝟐 𝟏𝟏𝟐𝟐 𝟏𝟏𝟒𝟒 𝟏𝟏𝟓𝟓 𝟗𝟗 𝟏𝟏𝟎𝟎

Example (8 minutes)

Example

A local car dealership is trying to sell all of the cars that are on the lot. Currently, there are 𝟓𝟓𝟐𝟐𝟓𝟓 cars on the lot, and the general manager estimates that they will consistently sell 𝟓𝟓𝟎𝟎 cars per week. Estimate how many weeks it will take for the number of cars on the lot to be less than 𝟕𝟕𝟓𝟓.

Write an inequality that can be used to find the number of full weeks, 𝒘𝒘, it will take for the number of cars to be less than 𝟕𝟕𝟓𝟓. Since 𝒘𝒘 is the number of full or complete weeks, 𝒘𝒘 = 𝟏𝟏 means at the end of week 1.

𝟓𝟓𝟐𝟐𝟓𝟓 − 𝟓𝟓𝟎𝟎𝒘𝒘 < 𝟕𝟕𝟓𝟓

Solve and graph the inequality.

𝟓𝟓𝟐𝟐𝟓𝟓 − 𝟓𝟓𝟎𝟎𝒘𝒘 < 𝟕𝟕𝟓𝟓

−𝟓𝟓𝟎𝟎𝒘𝒘 + 𝟓𝟓𝟐𝟐𝟓𝟓 − 𝟓𝟓𝟐𝟐𝟓𝟓 < 𝟕𝟕𝟓𝟓 − 𝟓𝟓𝟐𝟐𝟓𝟓 −𝟓𝟓𝟎𝟎𝒘𝒘 + 𝟎𝟎 < −𝟒𝟒𝟓𝟓𝟎𝟎

�−𝟏𝟏𝟓𝟓𝟎𝟎� (−𝟓𝟓𝟎𝟎𝒘𝒘) > �−

𝟏𝟏𝟓𝟓𝟎𝟎� (−𝟒𝟒𝟓𝟓𝟎𝟎)

𝒘𝒘 > 𝟗𝟗

Interpret the solution in the context of the problem.

The dealership can sell 𝟓𝟓𝟎𝟎 cars per week for more than 𝟗𝟗 weeks to have less than 𝟕𝟕𝟓𝟓 cars remaining on the lot.

Verify the solution.

𝒘𝒘 = 𝟗𝟗:

𝟓𝟓𝟐𝟐𝟓𝟓 − 𝟓𝟓𝟎𝟎𝒘𝒘 < 𝟕𝟕𝟓𝟓 𝟓𝟓𝟐𝟐𝟓𝟓 − 𝟓𝟓𝟎𝟎(𝟗𝟗) < 𝟕𝟕𝟓𝟓

𝟓𝟓𝟐𝟐𝟓𝟓 − 𝟒𝟒𝟓𝟓𝟎𝟎 < 𝟕𝟕𝟓𝟓 𝟕𝟕𝟓𝟓 < 𝟕𝟕𝟓𝟓

False

𝒘𝒘 = 𝟏𝟏𝟎𝟎:

𝟓𝟓𝟐𝟐𝟓𝟓 − 𝟓𝟓𝟎𝟎𝒘𝒘 < 𝟕𝟕𝟓𝟓 𝟓𝟓𝟐𝟐𝟓𝟓 − 𝟓𝟓𝟎𝟎(𝟏𝟏𝟎𝟎) < 𝟕𝟕𝟓𝟓

𝟓𝟓𝟐𝟐𝟓𝟓 − 𝟓𝟓𝟎𝟎𝟎𝟎 < 𝟕𝟕𝟓𝟓 𝟐𝟐𝟓𝟓 < 𝟕𝟕𝟓𝟓

True

Explain why 50𝑤𝑤 was subtracted from 525 and why the less than inequality was used.

Subtraction was used because the cars are being sold. Therefore, the inventory is being reduced. The less than inequality was used because the question asked for the number of cars remaining to be less than 75.

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7•3 Lesson 15

In one of the steps, the inequality was reversed. Why did this occur?

The inequality in the problem reversed because both sides were multiplied by a negative number.

Exercise 2 (Optional, 8 minutes)

Have students complete the exercise individually then compare answers with a partner.

Exercise 2

2. The cost of renting a car is $𝟐𝟐𝟓𝟓 per day plus a one-time fee of $𝟕𝟕𝟓𝟓.𝟓𝟓𝟎𝟎 for insurance. How many days can the car be rented if the total cost is to be no more than $𝟓𝟓𝟐𝟐𝟓𝟓?

a. Write an inequality to model the situation.

Let 𝒙𝒙 represent the number of days the car is rented.

𝟐𝟐𝟓𝟓𝒙𝒙 + 𝟕𝟕𝟓𝟓.𝟓𝟓𝟎𝟎 ≤ 𝟓𝟓𝟐𝟐𝟓𝟓

b. Solve and graph the inequality.

𝟐𝟐𝟓𝟓𝒙𝒙 + 𝟕𝟕𝟓𝟓.𝟓𝟓𝟎𝟎 ≤ 𝟓𝟓𝟐𝟐𝟓𝟓 𝟐𝟐𝟓𝟓𝒙𝒙 + 𝟕𝟕𝟓𝟓.𝟓𝟓𝟎𝟎 − 𝟕𝟕𝟓𝟓.𝟓𝟓𝟎𝟎 ≤ 𝟓𝟓𝟐𝟐𝟓𝟓 − 𝟕𝟕𝟓𝟓.𝟓𝟓𝟎𝟎

𝟐𝟐𝟓𝟓𝒙𝒙 + 𝟎𝟎 ≤ 𝟒𝟒𝟒𝟒𝟗𝟗.𝟓𝟓𝟎𝟎

�𝟏𝟏𝟐𝟐𝟓𝟓� (𝟐𝟐𝟓𝟓𝒙𝒙) ≤ �

𝟏𝟏𝟐𝟐𝟓𝟓� (𝟒𝟒𝟒𝟒𝟗𝟗.𝟓𝟓𝟎𝟎)

𝒙𝒙 ≤ 𝟏𝟏𝟕𝟕.𝟗𝟗𝟖𝟖

OR

𝟐𝟐𝟓𝟓𝒙𝒙 + 𝟕𝟕𝟓𝟓.𝟓𝟓𝟎𝟎 ≤ 𝟓𝟓𝟐𝟐𝟓𝟓 𝟐𝟐,𝟓𝟓𝟎𝟎𝟎𝟎𝒙𝒙 + 𝟕𝟕,𝟓𝟓𝟓𝟓𝟎𝟎 ≤ 𝟓𝟓𝟐𝟐,𝟓𝟓𝟎𝟎𝟎𝟎

𝟐𝟐,𝟓𝟓𝟎𝟎𝟎𝟎𝒙𝒙 + 𝟕𝟕,𝟓𝟓𝟓𝟓𝟎𝟎 − 𝟕𝟕,𝟓𝟓𝟓𝟓𝟎𝟎 ≤ 𝟓𝟓𝟐𝟐,𝟓𝟓𝟎𝟎𝟎𝟎 − 𝟕𝟕,𝟓𝟓𝟓𝟓𝟎𝟎

�𝟏𝟏

𝟐𝟐,𝟓𝟓𝟎𝟎𝟎𝟎� (𝟐𝟐,𝟓𝟓𝟎𝟎𝟎𝟎𝒙𝒙) ≤ �

𝟏𝟏𝟐𝟐,𝟓𝟓𝟎𝟎𝟎𝟎

� (𝟒𝟒𝟒𝟒,𝟗𝟗𝟓𝟓𝟎𝟎)

𝒙𝒙 ≤ 𝟏𝟏𝟕𝟕.𝟗𝟗𝟖𝟖

c. Interpret the solution in the context of the problem.

The car can be rented for 𝟏𝟏𝟕𝟕 days or fewer and stay within the amount of $𝟓𝟓𝟐𝟐𝟓𝟓. The number of days is an integer. The 18th day would put the cost over $𝟓𝟓𝟐𝟐𝟓𝟓, and since the fee is charged per day, the solution set includes whole numbers.

Game or Additional Exercises (12 minutes)

Make copies of the puzzle at the end of the lesson and cut the puzzle into 16 smaller squares. Mix up the pieces. Give each student a puzzle and tell him to put the pieces together to form a 4 × 4 square. When pieces are joined, the problem on one side must be attached to the answer on the other. All problems on the top, bottom, right, and left must line up to the correct graph of the solution. The puzzle, as it is shown, is the answer key.

𝟏𝟏𝟐𝟐 𝟏𝟏𝟒𝟒 𝟏𝟏𝟓𝟓 𝟏𝟏𝟔𝟔 𝟏𝟏𝟕𝟕 𝟏𝟏𝟖𝟖

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7•3 Lesson 15

Additional Exercises (in Lieu of the Game)

For each problem, write, solve, and graph the inequality, and interpret the solution within the context of the problem.

3. Mrs. Smith decides to buy three sweaters and a pair of jeans. She has $𝟏𝟏𝟐𝟐𝟎𝟎 in her wallet. If the price of the jeans is $𝟐𝟐𝟓𝟓, what is the highest possible price of a sweater, if each sweater is the same price?

Let 𝒘𝒘 represent the price of one sweater.

𝟐𝟐𝒘𝒘 + 𝟐𝟐𝟓𝟓 ≤ 𝟏𝟏𝟐𝟐𝟎𝟎 𝟐𝟐𝒘𝒘 + 𝟐𝟐𝟓𝟓 − 𝟐𝟐𝟓𝟓 ≤ 𝟏𝟏𝟐𝟐𝟎𝟎 − 𝟐𝟐𝟓𝟓

𝟐𝟐𝒘𝒘 + 𝟎𝟎 ≤ 𝟖𝟖𝟓𝟓

�𝟏𝟏𝟐𝟐� (𝟐𝟐𝒘𝒘) ≤ �

𝟏𝟏𝟐𝟐� (𝟖𝟖𝟓𝟓)

𝒘𝒘 ≤ 𝟐𝟐𝟖𝟖.𝟐𝟐𝟐𝟐

Graph:

Solution: The highest price Mrs. Smith can pay for a sweater and have enough money is $𝟐𝟐𝟖𝟖.𝟐𝟐𝟐𝟐.

4. The members of the Select Chorus agree to buy at least 𝟐𝟐𝟓𝟓𝟎𝟎 tickets for an outside concert. They buy 𝟐𝟐𝟎𝟎 fewer lawn tickets than balcony tickets. What is the least number of balcony tickets bought?

Let 𝒃𝒃 represent the number of balcony tickets.

Then 𝒃𝒃 − 𝟐𝟐𝟎𝟎 represents the number of lawn tickets.

𝒃𝒃 + 𝒃𝒃 − 𝟐𝟐𝟎𝟎 ≥ 𝟐𝟐𝟓𝟓𝟎𝟎 𝟐𝟐𝒃𝒃 − 𝟐𝟐𝟎𝟎 ≥ 𝟐𝟐𝟓𝟓𝟎𝟎

𝟐𝟐𝒃𝒃 − 𝟐𝟐𝟎𝟎 + 𝟐𝟐𝟎𝟎 ≥ 𝟐𝟐𝟓𝟓𝟎𝟎+ 𝟐𝟐𝟎𝟎 𝟐𝟐𝒃𝒃 + 𝟎𝟎 ≥ 𝟐𝟐𝟕𝟕𝟎𝟎

�𝟏𝟏𝟐𝟐� (𝟐𝟐𝒃𝒃) ≥ �

𝟏𝟏𝟐𝟐� (𝟐𝟐𝟕𝟕𝟎𝟎)

𝒃𝒃 ≥ 𝟏𝟏𝟐𝟐𝟓𝟓

Graph:

Solution: The least number of balcony tickets bought is 𝟏𝟏𝟐𝟐𝟓𝟓. The answers need to be integers.

𝟏𝟏𝟐𝟐𝟓𝟓 𝟏𝟏𝟐𝟐𝟔𝟔 𝟏𝟏𝟐𝟐𝟕𝟕 𝟏𝟏𝟐𝟐𝟖𝟖 𝟏𝟏𝟐𝟐𝟗𝟗 𝟏𝟏𝟐𝟐𝟎𝟎 𝟏𝟏𝟐𝟐𝟏𝟏 𝟏𝟏𝟐𝟐𝟐𝟐 𝟏𝟏𝟐𝟐𝟐𝟐 𝟏𝟏𝟐𝟐𝟖𝟖 𝟏𝟏𝟐𝟐𝟗𝟗 𝟏𝟏𝟒𝟒𝟎𝟎 𝟏𝟏𝟐𝟐𝟒𝟒 𝟏𝟏𝟐𝟐𝟓𝟓 𝟏𝟏𝟐𝟐𝟔𝟔 𝟏𝟏𝟐𝟐𝟕𝟕

𝟏𝟏𝟐𝟐 𝟏𝟏𝟐𝟐 𝟏𝟏𝟒𝟒 𝟏𝟏𝟓𝟓 𝟏𝟏𝟔𝟔 𝟏𝟏𝟕𝟕 𝟏𝟏𝟖𝟖 𝟏𝟏𝟗𝟗 𝟐𝟐𝟎𝟎 𝟐𝟐𝟓𝟓 𝟐𝟐𝟔𝟔 𝟐𝟐𝟕𝟕 𝟐𝟐𝟏𝟏 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟐𝟐𝟒𝟒 𝟐𝟐𝟖𝟖 𝟐𝟐𝟗𝟗 𝟐𝟐𝟎𝟎

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7•3 Lesson 15

𝟎𝟎 𝟏𝟏 𝟐𝟐 𝟐𝟐 𝟒𝟒 𝟓𝟓

5. Samuel needs $𝟐𝟐𝟗𝟗 to download some songs and movies on his MP3 player. His mother agrees to pay him $𝟔𝟔 an hour for raking leaves in addition to his $𝟓𝟓 weekly allowance. What is the minimum number of hours Samuel must work in one week to have enough money to purchase the songs and movies?

Let 𝒉𝒉 represent the number of hours Samuel rakes leaves.

𝟔𝟔𝒉𝒉+ 𝟓𝟓 ≥ 𝟐𝟐𝟗𝟗 𝟔𝟔𝒉𝒉 + 𝟓𝟓 − 𝟓𝟓 ≥ 𝟐𝟐𝟗𝟗 − 𝟓𝟓

𝟔𝟔𝒉𝒉+ 𝟎𝟎 ≥ 𝟐𝟐𝟒𝟒

�𝟏𝟏𝟔𝟔� (𝟔𝟔𝒉𝒉) ≥ �

𝟏𝟏𝟔𝟔� (𝟐𝟐𝟒𝟒)

𝒉𝒉 ≥ 𝟒𝟒

Graph:

Solution: Samuel needs to rake leaves at least 𝟒𝟒 hours to earn $𝟐𝟐𝟗𝟗. Any amount of time over 𝟒𝟒 hours will earn him extra money.

Closing (3 minutes)

Why do we use rays when graphing the solutions of an inequality on a number line?

When graphing the solution of an inequality on a number line, how do you determine what type of circle to use (open or closed)?

When graphing the solution of an inequality on a number line, how do you determine the direction of the arrow?


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7•3 Lesson 15

Name ___________________________________________________ Date____________________


Exit Ticket The junior high art club sells candles for a fundraiser. The first week of the fundraiser, the club sells 7 cases of candles. Each case contains 40 candles. The goal is to sell at least 13 cases. During the second week of the fundraiser, the club meets its goal. Write, solve, and graph an inequality that can be used to find the possible number of candles sold the second week.

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7•3 Lesson 15

𝟎𝟎 𝟏𝟏 𝟓𝟓 𝟕𝟕 𝟖𝟖 𝟗𝟗 𝟏𝟏𝟎𝟎 𝟐𝟐 𝟐𝟐 𝟒𝟒 𝟔𝟔


The junior high art club sells candles for a fundraiser. The first week of the fundraiser, the club sells 𝟕𝟕 cases of candles. Each case contains 𝟒𝟒𝟎𝟎 candles. The goal is to sell at least 𝟏𝟏𝟐𝟐 cases. During the second week of the fundraiser, the club meets its goal. Write, solve, and graph an inequality that can be used to find the minimum number of candles sold the second week.

Let 𝒏𝒏 represent the number candles sold the second week.

𝒏𝒏𝟒𝟒𝟎𝟎

+ 𝟕𝟕 ≥ 𝟏𝟏𝟐𝟐 𝒏𝒏𝟒𝟒𝟎𝟎

+ 𝟕𝟕 − 𝟕𝟕 ≥ 𝟏𝟏𝟐𝟐 − 𝟕𝟕 𝒏𝒏𝟒𝟒𝟎𝟎

≥ 𝟔𝟔

(𝟒𝟒𝟎𝟎) �𝒏𝒏𝟒𝟒𝟎𝟎� ≥ 𝟔𝟔(𝟒𝟒𝟎𝟎)

𝒏𝒏 ≥ 𝟐𝟐𝟒𝟒𝟎𝟎

The minimum number of candles sold the second week was 𝟐𝟐𝟒𝟒𝟎𝟎.

OR

Let 𝒏𝒏 represent the number of cases of candles sold the second week.

𝟒𝟒𝟎𝟎𝒏𝒏+ 𝟐𝟐𝟖𝟖𝟎𝟎 ≥ 𝟓𝟓𝟐𝟐𝟎𝟎 𝟒𝟒𝟎𝟎𝒏𝒏+ 𝟐𝟐𝟖𝟖𝟎𝟎 − 𝟐𝟐𝟖𝟖𝟎𝟎 ≥ 𝟓𝟓𝟐𝟐𝟎𝟎 − 𝟐𝟐𝟖𝟖𝟎𝟎

𝟒𝟒𝟎𝟎𝒏𝒏 + 𝟎𝟎 ≥ 𝟐𝟐𝟒𝟒𝟎𝟎

�𝟏𝟏𝟒𝟒𝟎𝟎� (𝟒𝟒𝟎𝟎𝒏𝒏) ≥ 𝟐𝟐𝟒𝟒𝟎𝟎�

𝟏𝟏𝟒𝟒𝟎𝟎�

𝒏𝒏 ≥ 𝟔𝟔

The minimum number of cases sold the second week was 𝟔𝟔. Since there are 𝟒𝟒𝟎𝟎 candles in each case, the minimum number of candles sold the second week would be (𝟒𝟒𝟎𝟎)(𝟔𝟔) = 𝟐𝟐𝟒𝟒𝟎𝟎.

𝟐𝟐𝟎𝟎𝟎𝟎 𝟐𝟐𝟏𝟏𝟎𝟎 𝟐𝟐𝟐𝟐𝟎𝟎 𝟐𝟐𝟐𝟐𝟎𝟎 𝟐𝟐𝟒𝟒𝟎𝟎 𝟐𝟐𝟓𝟓𝟎𝟎 𝟐𝟐𝟔𝟔𝟎𝟎

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7•3 Lesson 15


1. Ben has agreed to play fewer video games and spend more time studying. He has agreed to play less than 𝟏𝟏𝟎𝟎 hours

of video games each week. On Monday through Thursday, he plays video games for a total of 𝟓𝟓𝟏𝟏𝟐𝟐 hours. For the remaining 𝟐𝟐 days, he plays video games for the same amount of time each day. Find 𝒕𝒕, the amount of time he plays video games for each of the 𝟐𝟐 days. Graph your solution.

Let 𝒕𝒕 represent the time in hours spent playing video games.

𝟐𝟐𝒕𝒕+ 𝟓𝟓𝟏𝟏𝟐𝟐

< 𝟏𝟏𝟎𝟎

𝟐𝟐𝒕𝒕 + 𝟓𝟓𝟏𝟏𝟐𝟐− 𝟓𝟓

𝟏𝟏𝟐𝟐

< 𝟏𝟏𝟎𝟎 − 𝟓𝟓𝟏𝟏𝟐𝟐

𝟐𝟐𝒕𝒕+ 𝟎𝟎 < 𝟒𝟒𝟏𝟏𝟐𝟐

�𝟏𝟏𝟐𝟐� (𝟐𝟐𝒕𝒕) < �

𝟏𝟏𝟐𝟐� �𝟒𝟒

𝟏𝟏𝟐𝟐�

𝒕𝒕 < 𝟏𝟏.𝟓𝟓

Graph:

Ben plays less than 𝟏𝟏.𝟓𝟓 hours of video games each of the three days.

2. Gary’s contract states that he must work more than 𝟐𝟐𝟎𝟎 hours per week. The graph below represents the number of hours he can work in a week.

a. Write an algebraic inequality that represents the number of hours, 𝒉𝒉, Gary can work in a week.

𝒉𝒉 > 𝟐𝟐𝟎𝟎

b. Gary is paid $𝟏𝟏𝟓𝟓.𝟓𝟓𝟎𝟎 per hour in addition to a weekly salary of $𝟓𝟓𝟎𝟎. This week he wants to earn more than $𝟒𝟒𝟎𝟎𝟎𝟎. Write an inequality to represent this situation.

𝟏𝟏𝟓𝟓.𝟓𝟓𝟎𝟎𝒉𝒉 + 𝟓𝟓𝟎𝟎 > 𝟒𝟒𝟎𝟎𝟎𝟎

c. Solve and graph the solution from part (b). Round your answer to the nearest hour.

𝟏𝟏𝟓𝟓.𝟓𝟓𝟎𝟎𝒉𝒉 + 𝟓𝟓𝟎𝟎 − 𝟓𝟓𝟎𝟎 > 𝟒𝟒𝟎𝟎𝟎𝟎 − 𝟓𝟓𝟎𝟎

𝟏𝟏𝟓𝟓.𝟓𝟓𝟎𝟎𝒉𝒉 > 𝟐𝟐𝟓𝟓𝟎𝟎

�𝟏𝟏

𝟏𝟏𝟓𝟓.𝟓𝟓𝟎𝟎� (𝟏𝟏𝟓𝟓.𝟓𝟓𝟎𝟎𝒉𝒉) > 𝟐𝟐𝟓𝟓𝟎𝟎�

𝟏𝟏𝟏𝟏𝟓𝟓.𝟓𝟓𝟎𝟎

�

𝒉𝒉 > 𝟐𝟐𝟐𝟐.𝟓𝟓𝟖𝟖

Gary has to work 𝟐𝟐𝟐𝟐 or more hours to earn more than $𝟒𝟒𝟎𝟎𝟎𝟎.

𝟎𝟎 𝟎𝟎.𝟓𝟓 𝟏𝟏 𝟏𝟏.𝟓𝟓 𝟐𝟐 𝟐𝟐.𝟓𝟓 𝟐𝟐 𝟐𝟐.𝟓𝟓 𝟒𝟒 𝟒𝟒.𝟓𝟓 𝟓𝟓

𝟏𝟏𝟕𝟕 𝟏𝟏𝟖𝟖 𝟏𝟏𝟗𝟗 𝟐𝟐𝟎𝟎 𝟐𝟐𝟏𝟏 𝟐𝟐𝟐𝟐

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7•3 Lesson 15

𝟗𝟗𝟔𝟔 𝟗𝟗𝟕𝟕 𝟗𝟗𝟖𝟖 𝟗𝟗𝟗𝟗 𝟏𝟏𝟎𝟎𝟐𝟐 𝟏𝟏𝟎𝟎𝟐𝟐 𝟏𝟏𝟎𝟎𝟒𝟒 𝟏𝟏𝟎𝟎𝟎𝟎 𝟏𝟏𝟎𝟎𝟏𝟏

3. Sally’s bank account has $𝟔𝟔𝟓𝟓𝟎𝟎 in it. Every week, Sally withdraws $𝟓𝟓𝟎𝟎 to pay for her dog sitter. What is the maximum number of weeks that Sally can withdraw the money so there is at least $𝟕𝟕𝟓𝟓 remaining in the account? Write and solve an inequality to find the solution, and graph the solution on a number line.

Let 𝒘𝒘 represent the number of weeks Sally can withdraw the money.

𝟔𝟔𝟓𝟓𝟎𝟎 − 𝟓𝟓𝟎𝟎𝒘𝒘 ≥ 𝟕𝟕𝟓𝟓 𝟔𝟔𝟓𝟓𝟎𝟎 − 𝟓𝟓𝟎𝟎𝒘𝒘− 𝟔𝟔𝟓𝟓𝟎𝟎 ≥ 𝟕𝟕𝟓𝟓 − 𝟔𝟔𝟓𝟓𝟎𝟎

−𝟓𝟓𝟎𝟎𝒘𝒘 ≥ −𝟓𝟓𝟕𝟕𝟓𝟓

�𝟏𝟏

−𝟓𝟓𝟎𝟎� (−𝟓𝟓𝟎𝟎𝒘𝒘) ≥ �

𝟏𝟏−𝟓𝟓𝟎𝟎

� (−𝟓𝟓𝟕𝟕𝟓𝟓)

𝒘𝒘 ≤ 𝟏𝟏𝟏𝟏.𝟓𝟓

The maximum number of weeks Sally can withdraw the weekly dog sitter fee is 𝟏𝟏𝟏𝟏 weeks.

4. On a cruise ship, there are two options for an Internet connection. The first option is a fee of $𝟓𝟓 plus an additional $𝟎𝟎.𝟐𝟐𝟓𝟓 per minute. The second option costs $𝟓𝟓𝟎𝟎 for an unlimited number of minutes. For how many minutes, 𝒎𝒎, is the first option cheaper than the second option? Graph the solution.

Let 𝒎𝒎 represent the number of minutes of Internet connection.

𝟓𝟓 + 𝟎𝟎.𝟐𝟐𝟓𝟓𝒎𝒎 < 𝟓𝟓𝟎𝟎

𝟓𝟓 + 𝟎𝟎.𝟐𝟐𝟓𝟓𝒎𝒎− 𝟓𝟓 < 𝟓𝟓𝟎𝟎 − 𝟓𝟓

𝟎𝟎.𝟐𝟐𝟓𝟓𝒎𝒎+ 𝟎𝟎 < 𝟒𝟒𝟓𝟓

�𝟏𝟏

𝟎𝟎.𝟐𝟐𝟓𝟓� (𝟎𝟎.𝟐𝟐𝟓𝟓𝒎𝒎) < �

𝟏𝟏𝟎𝟎.𝟐𝟐𝟓𝟓

� (𝟒𝟒𝟓𝟓)

𝒎𝒎 < 𝟏𝟏𝟖𝟖𝟎𝟎

If there are less than 𝟏𝟏𝟖𝟖𝟎𝟎 minutes, or 𝟐𝟐 hours, used on the Internet, then the first option would be cheaper. If 𝟏𝟏𝟖𝟖𝟎𝟎 minutes or more are planned, then the second option is more economical.

5. The length of a rectangle is 𝟏𝟏𝟎𝟎𝟎𝟎 centimeters, and its perimeter is greater than 𝟒𝟒𝟎𝟎𝟎𝟎 centimeters. Henry writes an inequality and graphs the solution below to find the width of the rectangle. Is he correct? If yes, write and solve the inequality to represent the problem and graph. If no, explain the error(s) Henry made.

Henry’s graph is incorrect. The inequality should be 𝟐𝟐(𝟏𝟏𝟎𝟎𝟎𝟎) + 𝟐𝟐𝒘𝒘 > 𝟒𝟒𝟎𝟎𝟎𝟎. When you solve the inequality, you get 𝒘𝒘 > 𝟏𝟏𝟎𝟎𝟎𝟎. The circle on 𝟏𝟏𝟎𝟎𝟎𝟎 on the number line is correct; however, the circle should be an open circle since the perimeter is not equal to 𝟒𝟒𝟎𝟎𝟎𝟎. Also, the arrow should be pointing in the opposite direction because the perimeter is greater than 𝟒𝟒𝟎𝟎𝟎𝟎, which means the width is greater than 𝟏𝟏𝟎𝟎𝟎𝟎. The given graph indicates an inequality of less than or equal to.

𝟏𝟏𝟕𝟕𝟓𝟓 𝟏𝟏𝟕𝟕𝟔𝟔 𝟏𝟏𝟕𝟕𝟕𝟕 𝟏𝟏𝟕𝟕𝟖𝟖 𝟏𝟏𝟖𝟖𝟏𝟏 𝟏𝟏𝟖𝟖𝟐𝟐 𝟏𝟏𝟖𝟖𝟐𝟐 𝟏𝟏𝟕𝟕𝟗𝟗 𝟏𝟏𝟖𝟖𝟎𝟎

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7•3 Lesson 15

Inequalities Progression of Exercises

Determine the value(s) of the variable.

Set 1

1. 𝑝𝑝 + 1 > 8 𝒙𝒙 > 𝟕𝟕

2. 𝑝𝑝 + 3 > 8 𝒙𝒙 > 𝟓𝟓

3. 𝑝𝑝 + 10 > 8 𝒙𝒙 > −𝟐𝟐

4. 𝑝𝑝 − 2 > 3 𝒙𝒙 > 𝟓𝟓

5. 𝑝𝑝 − 4 > 3 𝒙𝒙 > 𝟕𝟕

Set 2

1. 3𝑝𝑝 ≤ 15 𝒙𝒙 ≤ 𝟓𝟓

2. 3𝑝𝑝 ≤ 21 𝒙𝒙 ≤ 𝟕𝟕

3. −𝑝𝑝 ≤ 4 𝒙𝒙 ≥ −𝟒𝟒

4. −2𝑝𝑝 ≤ 4

𝒙𝒙 ≥ −𝟐𝟐

5. −𝑝𝑝 ≤ −4

𝒙𝒙 ≥ 𝟒𝟒

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7•3 Lesson 15

Set 3

1. 12𝑝𝑝 < 1

𝒙𝒙 < 𝟐𝟐

2. 12𝑝𝑝 < 3

𝒙𝒙 < 𝟔𝟔

3. − 15 𝑝𝑝 < 2

𝒙𝒙 > −𝟏𝟏𝟎𝟎

4. − 25 𝑝𝑝 < 2

𝒙𝒙 > −𝟓𝟓

5. − 35 𝑝𝑝 < 3

𝒙𝒙 > −𝟓𝟓

Set 4

1. 2𝑝𝑝 + 4 ≥ 8 𝒙𝒙 ≥ 𝟐𝟐

2. 2𝑝𝑝 − 3 ≥ 5

𝒙𝒙 ≥ 𝟒𝟒

3. −2𝑝𝑝 + 1 ≥ 7

𝒙𝒙 ≤ −𝟐𝟐

4. −3𝑝𝑝 + 1 ≥ −8

𝒙𝒙 ≤ 𝟐𝟐

5. −3𝑝𝑝 − 5 ≥ 10

𝒙𝒙 ≤ −𝟓𝟓

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7•3 Lesson 15

Set 5

1. 2𝑝𝑝 − 0.5 > 5.5 𝒙𝒙 > 𝟐𝟐

2. 3𝑝𝑝 + 1.5 > 4.5 𝒙𝒙 > 𝟐𝟐

3. 5𝑝𝑝 − 3 > 4.5

𝒙𝒙 > 𝟏𝟏.𝟓𝟓

4. −5𝑝𝑝 + 2 > 8.5

𝒙𝒙 < −𝟏𝟏.𝟐𝟐

5. −9𝑝𝑝 − 3.5 > 1

𝒙𝒙 < −𝟎𝟎.𝟓𝟓

Set 6

1. 2(𝑝𝑝 + 3) ≤ 4 𝒙𝒙 ≤ −𝟏𝟏

2. 3(𝑝𝑝 + 3) ≤ 6

𝒙𝒙 ≤ −𝟏𝟏

3. 4(𝑝𝑝 + 3) ≤ 8

𝒙𝒙 ≤ −𝟏𝟏

4. −5(𝑝𝑝 − 3) ≤ −10

𝒙𝒙 ≥ 𝟓𝟓

5. −2(𝑝𝑝 + 3) ≤ 8

𝒙𝒙 ≥ −𝟕𝟕

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7•3 Lesson 15

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7•3 Mid-Module Assessment Task

Name Date Use the expression below to answer parts (a) and (b).

4𝑥𝑥 − 3(𝑥𝑥 − 2𝑦𝑦) +12

(6𝑥𝑥 − 8𝑦𝑦)

a. Write an equivalent expression in standard form, and collect like terms.

b. Express the answer from part (a) as an equivalent expression in factored form.

2. Use the information to solve the problems below.

a. The longest side of a triangle is six more units than the shortest side. The third side is twice the length of the shortest side. If the perimeter of the triangle is 25 units, write and solve an equation to find the lengths of all three sides of the triangle.

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b. The length of a rectangle is (𝑥𝑥 + 3) inches long, and the width is 3 25 inches. If the area is 15 3

10 square inches, write and solve an equation to find the length of the rectangle.

3. A picture 10 1

4 feet long is to be centered on a wall that is 14 12 feet long. How much space is there from

the edge of the wall to the picture?

a. Solve the problem arithmetically.

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b. Solve the problem algebraically. c. Compare the approaches used in parts (a) and (b). Explain how they are similar.

4. In August, Cory begins school shopping for his triplet daughters. a. One day, he bought 10 pairs of socks for $2.50 each and 3 pairs of shoes for 𝑑𝑑 dollars each. He

spent a total of $135.97. Write and solve an equation to find the cost of one pair of shoes.

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b. The following day Cory returned to the store to purchase some more socks. He had $40 to spend.

When he arrived at the store, the shoes were on sale for 13

off. What is the greatest amount of pairs of socks Cory can purchase if he purchases another pair of shoes in addition to the socks?

5. Ben wants to have his birthday at the bowling alley with a few of his friends, but he can spend no more

than $80. The bowling alley charges a flat fee of $45 for a private party and $5.50 per person for shoe rentals and unlimited bowling. a. Write an inequality that represents the total cost of Ben’s birthday for 𝑝𝑝 people given his budget.

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b. How many people can Ben pay for (including himself) while staying within the limitations of his budget?

c. Graph the solution of the inequality from part (a).

6. Jenny invited Gianna to go watch a movie with her family. The movie theater charges one rate for 3D admission and a different rate for regular admission. Jenny and Gianna decided to watch the newest movie in 3D. Jenny’s mother, father, and grandfather accompanied Jenny’s little brother to the regular admission movie. a. Write an expression for the total cost of the tickets. Define the variables.

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b. The cost of the 3D ticket was double the cost of the regular admission ticket. Write an equation to represent the relationship between the two types of tickets.

c. The family purchased refreshments and spent a total of $18.50. If the total amount of money spent

on tickets and refreshments was $94.50, use an equation to find the cost of one regular admission ticket.

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7. The three lines shown in the diagram below intersect at the same point. The measures of some of the angles in degrees are given as 3(𝑥𝑥 − 2)°, �3

5𝑦𝑦� °, 12°, 42°.

a. Write and solve an equation that can be used to find the value of 𝑥𝑥.

b. Write and solve an equation that can be used to find the value of 𝑦𝑦.

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A Progression Toward Mastery

Assessment Task Item

STEP 1 Missing or incorrect answer and little evidence of reasoning or application of mathematics to solve the problem

STEP 2 Missing or incorrect answer but evidence of some reasoning or application of mathematics to solve the problem

STEP 3 A correct answer with some evidence of reasoning or application of mathematics to solve the problem, or an incorrect answer with substantial evidence of solid reasoning or application of mathematics to solve the problem

STEP 4 A correct answer supported by substantial evidence of solid reasoning or application of mathematics to solve the problem

1

a 7.EE.A.1

Student demonstrates a limited understanding of writing the expression in standard form. OR Student makes a conceptual error, such as dropping the parentheses or adding instead of multiplying.

Student makes two or more computational errors.

Student demonstrates a solid understanding but makes one computational error and completes the question by writing a correct equivalent expression in standard form. OR Student answers correctly, but no further correct work is shown.

Student writes the expression correctly in standard form, 4𝑥𝑥 + 2𝑦𝑦. Student shows appropriate work, such as using the distributive property and collecting like terms.

b

7.EE.A.1

Student demonstrates a limited understanding of writing the expression in factored form.

Student writes the expression correctly in factored form but has part (a) incorrect.

Student demonstrates a solid understanding by writing a correct equivalent expression in factored form based on the answer from part (a) but makes one computational error.

Student uses the correct expression from part (a). Student writes an equivalent expression in factored form.

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2 a

7.EE.B.3 7.EE.B.4a

Student demonstrates a limited understanding of perimeter by finding three sides of a triangle whose sum is 25, but the sides are incorrect and do not satisfy the given conditions. For example, a student says the sides are 4, 10, and 11 because they add up to 25 but does not satisfy the given conditions.

Student makes a conceptual error and one computational error. OR Student makes a conceptual error but writes an equation of equal difficulty and solves it correctly but does not find the lengths of the sides of the triangle.

Student demonstrates a solid understanding and finishes the problem correctly but makes one computational error with a value still resulting in a fractional value. OR Student sets up and solves an equation correctly but does not substitute the value back into the expressions to determine the actual side lengths. OR Student defines the sides correctly and sets up an equation but makes one error in solving the equation and finding the corresponding side lengths.

Student correctly defines the variable, sets up an equation to represent the perimeter, such as 2𝑥𝑥 + 𝑥𝑥 + 𝑥𝑥 + 6 = 25, solves the equation

correctly, 𝑥𝑥 = 4 34 , and

determines the lengths of the three sides to be

4 34, 9 1

2, and 10 34.

OR Student finds the correct lengths of the sides of a

triangle to be 4 34, 9 1

2,

and 10 34, using

arithmetic or a tape diagram, showing appropriate and correct work.

b

7.EE.B.3 7.EE.B.4a

Student makes a conceptual error, such as finding the perimeter, and makes two or more computational errors, without finding the length.

Student writes a correct equation demonstrating area, but no further correct work is shown. OR Student makes a conceptual error, such as finding perimeter instead of area, and makes one computational error solving the equation but finds the appropriate length.

Student demonstrates the concept of area and finds the appropriate length based on the answer obtained but makes one or two computational errors. OR Student finds the correct value of 𝑥𝑥 but does not determine the length of the rectangle. OR Student makes a conceptual error such as adding to find the perimeter instead of multiplying for the area. For example, student sets up the following equation of equal

difficulty: 3 25 + 3 2

5 +

𝑥𝑥 + 3 + 𝑥𝑥 + 3 = 15 310;

student solves the equation correctly, 𝑥𝑥 = 1 1

4; and finds the correct length according to the answer obtained.

Student correctly defines the variable, sets up an equation to represent the area, such as

3 25 (𝑥𝑥 + 3) = 15 3

10, solves the equation

correctly, 𝑥𝑥 = 1 12, and

determines the length to

be 4 12 inches.

Student uses arithmetic or a tape diagram, and shows appropriate and correct work.

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3

a

7.EE.B.3

Student shows a limited understanding but makes a conceptual error, such as finding half of the sum of the lengths.

Student finds the correct difference between the length of the wall and the length of the picture

as 4 14 inches, but no

further correct work is shown.

Student shows that half the difference must be found but makes a computational error.

Student demonstrates understanding by finding half of the difference between the length of the wall and the length of the picture and shows appropriate work, possibly using a tape diagram, to obtain an

answer of 2 18 inches.

b

7.EE.B.4a

Student demonstrates a limited understanding of writing an equation to demonstrate the situation, but very little correct work is shown.

Student makes a conceptual error writing the equation, such as

𝑥𝑥 + 10 14 = 14 1

2. OR Student makes a conceptual error in solving the equation. OR Student makes two or more computational errors in solving the equation.

Student sets up a correct equation but makes one computational error.

Student correctly defines a variable, sets up an equation such as

𝑥𝑥 + 10 14 + 𝑥𝑥 = 14 1

2, and finds the correct

value of 2 18 inches.

c

7.EE.B.3 7.EE.B.4a

Student demonstrates little or no understanding between an arithmetic and an algebraic approach.

Student recognizes the solutions are the same. OR Student recognizes the operations are the same.

Student recognizes the solutions are the same. AND Student recognizes the operations are the same.

Student recognizes the solutions, the operations, and the order of the operations are the same between an arithmetic and an algebraic approach.

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4 a

7.EE.A.2 7.EE.B.4a

Student demonstrates a limited understanding by writing an incorrect equation and solving it incorrectly.

Student makes a conceptual error in solving the equation, such as subtracting by 3 instead of multiplying by 13

.

OR Student makes a conceptual error in writing the equation, such as 10 + 2.50 +3𝑑𝑑 = 135.97 or 2.50 + 3𝑑𝑑 = 135.97, but further work is solved correctly. OR Student writes a correct equation, but no further correct work is shown. OR Student finds the correct answer without writing an equation, such as using a tape diagram or arithmetic.

Student sets up a correct equation but makes one computational error. OR Student finds the correct value of the variable but does not state the cost of one pair of shoes. OR Student sets up a wrong equation of equal difficulty by writing a number from the problem incorrectly, but all further work is correct.

Student clearly defines the variable, writes a correct equation, such as 10(2.50) + 3𝑑𝑑 =135.97, and finds the correct cost of one pair of shoes as $36.99.

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b

7.EE.A.2 7.EE.B.4b

Student demonstrates a limited understanding of discount price, inequality, and solution of inequality. OR Student finds the correct new price for the shoes of $24.66, but no further correct work is shown. OR Student writes an inequality representing the total cost as an inequality, but no further correct work is shown, nor is the new price for the shoes found correctly.

Student makes a conceptual error such as not finding the new discount price of the shoes. OR Student makes a conceptual error in writing or solving the inequality. OR Student makes two or more computational or rounding errors.

Student demonstrates a solid understanding but makes one computational error. OR Student calculates the discount on the shoes incorrectly but finishes the remaining problem correctly. For example, student uses the discount amount, $12.33, as the new price of shoes but writes a correct inequality and solution of $11 based on the discount amount. OR Student determines the correct discount price for the shoes but uses ≥ instead of ≤. OR Student determines the correct discount price for the shoes, writes and solves the inequality correctly, but does not round or rounds incorrectly.

Student determines the correct new price for the

shoes including the 13

off

as $24.66, writes a correct inequality, 2.50𝑑𝑑 + 24.66 ≤ 40, solves the inequality correctly, 𝑑𝑑 ≤ 6.136, and determines by rounding correctly that the amount of socks that could be purchased is 6 pairs.

5 a

7.EE.B.4b

Student demonstrates little or no understanding of writing inequalities.

Student writes an inequality but makes two of the following errors: She writes the incorrect inequality sign, the variable is in the wrong location, or she writes a subtraction sign instead of an addition sign.

Student writes an inequality but makes one of the following errors: He writes the incorrect inequality sign, the variable is in the wrong location, or he writes a subtraction sign instead of an addition sign.

Student writes a correct inequality, 45 + 5.50𝑝𝑝 ≤ 80, to represent the situation.

b

7.EE.B.4b

Student demonstrates a limited understanding of solving the inequality and interpreting the solution.

Student makes a conceptual error in solving the inequality. OR Student makes two or more computational or rounding errors.

Student demonstrates a solid understanding of solving the inequality but makes one computational or one rounding error.

Student solves the inequality written from part (a) correctly,

𝑝𝑝 ≤ 6 411, and

determines the correct number of people by rounding correctly to 6.

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c

7.EE.B.4b

Student demonstrates a limited understanding of graphing inequalities by only plotting the point correctly.

Student graphs the inequality but makes two or more errors. OR Student makes a conceptual error such as graphing on a coordinate plane instead of a number line.

Student demonstrates a solid understanding of graphing an inequality but makes one error such as an open circle, an incorrect scale, a circle placed in the wrong area, or an arrow drawn in the wrong direction.

Student correctly graphs the solution of the inequality from part (b). An appropriate scale is provided, clearly showing 6 and 7. A closed circle is shown at 6, and an arrow is pointing to the left.

6 a

7.EE.B.4

Student demonstrates a limited understanding, such as indicating a sum, but the variables are not clearly defined, and the expression is left without collecting like terms.

Student makes a conceptual error such as finding the difference of all the costs of admissions. OR Student does not write an expression to find the total cost. Instead, the student leaves each admission as a separate expression, 2𝑑𝑑 and 4𝑟𝑟.

Student writes a correct expression but does not define the variables or does not define them correctly, such as 𝑑𝑑 represents the 3D admission and 𝑟𝑟 represents regular. The variables need to specifically indicate the cost. OR Student makes one computational error in collecting like terms. OR Student clearly defines the variables correctly but makes one mistake in the expression, such as leaving out one person.

Student clearly defines the variables and writes an expression such as 2𝑑𝑑 + 4𝑟𝑟, with appropriate work shown. The definition of the variables must indicate the cost of each admission.

b

7.EE.B.4a

Student demonstrates no understanding of writing equations.

Student writes an equation but makes two errors: He uses the incorrect coefficient, and the variable is in the wrong location.

Student writes an equation but makes one of the following errors: She uses the incorrect coefficient or the variable is in the wrong location.

An expression is written, such as 𝑑𝑑 = 2𝑟𝑟 or

𝑟𝑟 = 12𝑑𝑑, to demonstrate

that the cost of 3D admission is double, or two times, the cost of a regular admission ticket.

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c

7.EE.B.4a

Student demonstrates a limited understanding of writing an equation and solving the equation.

Student writes a correct equation but no further correct work is shown. OR Student makes a conceptual error, such as solving the equation but disregarding the two variables and making them one, such as 6𝑑𝑑. OR Student writes a correct equation but makes two or more computational errors.

Student writes a correct equation but makes one computational error. OR Student solves the equation correctly but does not indicate the final cost of admission as $9.50.

Student writes a correct equation, such as 2𝑑𝑑 + 4𝑟𝑟 + 18.50 =94.50, and solves it correctly by substituting 2𝑟𝑟 for 𝑑𝑑, resulting in 𝑑𝑑 = 9.5, and writes the correct answer of the cost of regular admission, $9.50.

7 a

7.G.B.5

Student demonstrates a limited understanding of vertical angle relationships by writing an equation such as 3(𝑥𝑥 − 2) = 12, but no further work is shown, or the work shown is incorrect.

Student writes the correct equation, but no further correct work is shown, or two or more computational errors are made solving the equation. OR Student makes a conceptual error such as adding the angles to equal 180, and all further work shown is correct. OR Student makes a conceptual error writing the equation, such as 3(𝑥𝑥 − 2) = 12, but solves the equation correctly, getting 𝑥𝑥 = 6, and all work is shown. OR Student writes a correct equation with two variables, 3(𝑥𝑥 − 2) + 3

4𝑦𝑦 + 12 =180, but no further work is shown.

Student writes the correct equation representing the vertical angle relationship but makes one computational error in solving the equation.

Student correctly recognizes the vertical angle relationship, writes the equation 3(𝑥𝑥 − 2) = 42, and solves the equation correctly, showing all work, and getting a value of 16 for 𝑥𝑥.

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b

7.G.B.5

Student demonstrates a limited understanding of supplementary angles adding up to equal 180, but the wrong angles are used, and no other correct work is shown.

Student writes the correct equation but makes two or more computational errors or a conceptual error in solving the equation. OR Student writes a correct equation with two variables, 3(𝑥𝑥 − 2) + 3

4𝑦𝑦 + 12 =180, but no further work is shown.

Student writes the correct equation but makes one computational error in solving.

Student writes a correct equation demonstrating the supplementary angles, 35𝑦𝑦 + 12 + 42 = 180,

solves the equation correctly showing all work, and arrives at 𝑦𝑦 = 210.

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Name Date

1. Use the following expression below to answer parts (a) and (b).

4𝑥𝑥 − 3(𝑥𝑥 − 2𝑦𝑦) +12

(6𝑥𝑥 − 8𝑦𝑦)

a. Write an equivalent expression in standard form, and collect like terms.

b. Express the answer from part (a) as an equivalent expression in factored form.

2. Use the following information to solve the problems below. a. The longest side of a triangle is six more units than the shortest side. The third side is twice the

length of the shortest side. If the perimeter of the triangle is 25 units, write and solve an equation to find the lengths of all three sides of the triangle.

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b. The length of a rectangle is (𝑥𝑥 + 3) inches long, and the width is 3 25 inches. If the area is 15 3

10 square inches, write and solve an equation to find the length of the rectangle.

3. A picture 10 14 feet long is to be centered on a wall that is 14 1

2 feet long. How much space is there from the edge of the wall to the picture? a. Solve the problem arithmetically.

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b. Solve the problem algebraically.

c. Compare the approaches used in parts (a) and (b). Explain how they are similar.

4. In August, Cory begins school shopping for his triplet daughters. a. One day, he bought 10 pairs of socks for $2.50 each and 3 pairs of shoes for 𝑑𝑑 dollars each. He

spent a total of $135.97. Write and solve an equation to find the cost of one pair of shoes.

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b. The following day Cory returned to the store to purchase some more socks. He had $40 to spend.

When he arrived at the store, the shoes were on sale for 13

off. What is the greatest amount of pairs of socks Cory can purchase if he purchases another pair of shoes in addition to the socks?

5. Ben wants to have his birthday at the bowling alley with a few of his friends, but he can spend no more than $80. The bowling alley charges a flat fee of $45 for a private party and $5.50 per person for shoe rentals and unlimited bowling. a. Write an inequality that represents the total cost of Ben’s birthday for 𝑝𝑝 people given his budget.

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7•3Mid-Module Assessment Task

b. How many people can Ben pay for (including himself) while staying within the limitations of his budget?

c. Graph the solution of the inequality from part (a).

6. Jenny invited Gianna to go watch a movie with her family. The movie theater charges one rate for 3D admission and a different rate for regular admission. Jenny and Gianna decided to watch the newest movie in 3D. Jenny’s mother, father, and grandfather accompanied Jenny’s little brother to the regular admission movie. a. Write an expression for the total cost of the tickets. Define the variables.

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b. The cost of the 3D ticket was double the cost of the regular admission. Write an equation to represent the relationship between the two types of tickets.

c. The family purchased refreshments and spent a total of $18.50. If the total amount of money spent on tickets and refreshments was $94.50, use an equation to find the cost of one regular admission ticket.

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7. The three lines shown in the diagram below intersect at the same point. The measures of some of the angles in degrees are given as 3(𝑥𝑥 − 2)°, �3

5𝑦𝑦� °, 12°, 42°.

a. Write and solve an equation that can be used to find the value of 𝑥𝑥.

b. Write and solve an equation that can be used to find the value of 𝑦𝑦.

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Topic C: Use Equations and Inequalities to Solve Geometry Problems


7 G R A D E

Mathematics Curriculum Topic C

Use Equations and Inequalities to Solve Geometry Problems

7.G.B.4, 7.G.B.6

Focus Standards: 7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle.

7.G.B.6 Solve real-world and mathematical problems involving area, volume and surface area of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms.


Lesson 16: The Most Famous Ratio of All (M)1

Lesson 17: The Area of a Circle (E)

Lesson 18: More Problems on Area and Circumference (P)

Lesson 19: Unknown Area Problems on the Coordinate Plane (P)

Lesson 20: Composite Area Problems (P)

Lessons 21–22: Surface Area (P)

Lessons 23–24: The Volume of a Right Prism (E)

Lessons 25–26: Volume and Surface Area (P)

Topic C begins with students discovering the greatest ratio of all, pi. In Lesson 16, students use a compass to construct a circle and extend their understanding of angles and arcs from earlier grades to develop the definition of a circle through exploration. A whole-group activity follows, in which a wheel, chalk, and string are used to physically model the ratio of a circle’s circumference to its diameter. Through this activity, students conceptualize pi as a number whose value is a little more than 3. The lesson continues to examine this relationship between a circle’s circumference and diameter, as students understand pi to be a constant, which can be represented using approximations.


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Topic C: Use Equations and Inequalities to Solve Geometry Problems

7•3Topic C

Students see the usefulness of approximations such as 227

and 3.14 to efficiently solve problems related to the circumference of circles and semicircles. Students continue examining circles in Lesson 17, as they discover what happens if they cut a circle of radius length 𝑟𝑟 into equivalent-sized sectors and rearrange them to resemble a rectangle. Applying what they know about the area of a rectangle, students examine the dimensions to derive a formula for the area of the circle (7.G.B.4). They use this formula, 𝐴𝐴 = 𝜋𝜋𝑟𝑟2, to solve problems with circles. In Lesson 18, students consider how to adapt the area and circumference formulas to examine interesting problems involving quarter circle and semicircle regions. Students analyze figures to determine composite area in Lesson 19 and 20 by composing and decomposing into familiar shapes. They use the coordinate plane as a tool to determine the length and area of figures with vertices at grid points.

This topic concludes as students apply their knowledge of plane figures to find the surface area and volume of three-dimensional figures. In Lessons 21 and 22, students use polyhedron nets to understand surface area as the sum of the area of the lateral faces and the area of the base(s) for figures composed of triangles and quadrilaterals. In Lessons 23 and 24, students recognize the volume of a right prism to be the area of the base times the height and compute volumes of right prisms involving fractional values for length (7.G.B.6). In the last two lessons, students solidify their understanding of two- and three-dimensional objects as they solve real-world and mathematical problems involving area, volume, and surface area.

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7•3 Lesson 16

Lesson 16: The Most Famous Ratio of All


Student Outcomes

Students develop the definition of a circle using diameter and radius. Students know that the distance around a circle is called the circumference and discover that the ratio of the

circumference to the diameter of a circle is a special number called pi, written 𝜋𝜋.

Students know the formula for the circumference 𝐶𝐶 of a circle, of diameter 𝑑𝑑, and radius 𝑟𝑟. They use scale models to derive these formulas.

Students use 227

and 3.14 as estimates for 𝜋𝜋 and informally show that 𝜋𝜋 is slightly greater than 3.

Lesson Notes Although students were introduced to circles in kindergarten and worked with angles and arcs measures in Grades 4 and 5, they have not examined a precise definition of a circle. This lesson combines the definition of a circle with the application of constructions with a compass and straightedge to examine the ideas associated with circles and circular regions.

Classwork


Materials: Each student has a compass and metric ruler.

Opening Exercise

a. Using a compass, draw a circle like the picture to the right.

𝑪𝑪 is the center of the circle. The distance between 𝑪𝑪 and 𝑩𝑩 is the radius of the circle.

b. Write your own definition for the term circle.

Student responses will vary. Many might say, “It is round.” “It is curved.” “It has an infinite number of sides.” “The points are always the same distance from the center.” Analyze their definitions, showing how other figures such as ovals are also “round” or “curved.” Ask them what is special about the compass they used. (Answer: The distance between the spike and the pencil is fixed when drawing the circle.) Let them try defining a circle again with this new knowledge.

Present the following information about a circle.

CIRCLE: Given a point 𝑂𝑂 in the plane and a number 𝑟𝑟 > 0, the circle with center 𝑂𝑂 and radius 𝑟𝑟 is the set of all points in the plane whose distance from the point 𝑂𝑂 is equal to 𝑟𝑟.

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7•3 Lesson 16


What does the distance between the spike and the pencil on a compass represent in the definition above?

The radius 𝑟𝑟

What does the spike of the compass represent in the definition above?

The center 𝐶𝐶

What does the image drawn by the pencil represent in the definition above? The set of all points

c. Extend segment 𝑪𝑪𝑩𝑩 to a segment 𝑨𝑨𝑩𝑩, where 𝑨𝑨 is also a point on the circle.

The length of the segment 𝑨𝑨𝑩𝑩 is called the diameter of the circle.

d. The diameter is twice, or 𝟐𝟐 times, as long as the radius.

After each student measures and finds that the diameter is twice as long as the radius, display several student examples of different-sized circles to the class. Did everyone get a measure that was twice as long? Ask if a student can use the definition of a circle to explain why the diameter must be twice as long.

e. Measure the radius and diameter of each circle. The center of each circle is labeled 𝑪𝑪.

𝑪𝑪𝑩𝑩 = 𝟏𝟏.𝟓𝟓 𝐜𝐜𝐜𝐜, 𝑨𝑨𝑩𝑩 = 𝟑𝟑 𝐜𝐜𝐜𝐜, 𝑪𝑪𝑪𝑪 = 𝟐𝟐 𝐜𝐜𝐜𝐜, 𝑬𝑬𝑪𝑪 = 𝟒𝟒 𝐜𝐜𝐜𝐜

The radius of the largest circle is 𝟑𝟑 𝐜𝐜𝐜𝐜. The diameter is 𝟔𝟔 𝐜𝐜𝐜𝐜.

f. Draw a circle of radius 𝟔𝟔 𝐜𝐜𝐜𝐜.

MP.3

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7•3 Lesson 16


Part (f) may not be as easy as it seems. Let students grapple with how to measure 6 cm with a compass. One difficulty they might encounter is trying to measure 6 cm by putting the spike of the compass on the edge of the ruler (i.e., the 0 cm mark). Suggest either of the following: (1) Measure the compass from the 1 cm mark to the 7 cm mark, or (2) Mark two points 6 cm apart on the paper first; then, use one point as the center.

Mathematical Modeling Exercise (15 minutes)

Materials: a bicycle wheel (as large as possible), tape or chalk, a length of string long enough to measure the circumference of the bike wheel

Activity: Invite the entire class to come up to the front of the room to measure a length of string that is the same length as the distance around the bicycle wheel. Give them the tape or chalk and string, but do not tell them how to use these materials to measure the circumference, at least not yet. The goal is to set up several “ah-ha” moments for students. Give them time to try to wrap the string around the bicycle wheel. They will quickly find that this way of trying to measure the circumference is unproductive (the string will pop off). Lead them to the following steps for measuring the circumference, even if they do succeed with wrapping the string:

1. Mark a point on the wheel with a piece of masking tape or chalk.

2. Mark a starting point on the floor, align it with the mark on the wheel, and carefully roll the wheel so that it rolls one complete revolution.

3. Mark the endpoint on the floor with a piece of masking tape or chalk.

Dramatically walk from the beginning mark to the ending mark on the floor, declaring, “The length between these two marks is called the circumference of the wheel; it is the distance around the wheel. We can now easily measure that distance with string.” First, ask two students to measure a length of string using the marks; then, ask them to hold up the string directly above the marks in front of the rest of the class. Students are ready for the next “ah-ha” moment.

Why is this new way of measuring the string better than trying to wrap the string around the wheel? (Because it leads to an accurate measurement of the circumference.)

The circumference of any circle is always the same multiple of the diameter. Mathematicians call this number pi. It is one of the few numbers that is so special it has its own name. Let’s see if we can estimate the value of pi.

Take the wheel and carefully measure three diameter lengths using the wheel itself, as in the picture below.

Mark the three diameter lengths on the rope with a marker. Then, have students wrap the rope around the wheel itself.

If the circumference was measured carefully, students see that the string is three wheel diameters plus a little bit extra at the end. Have students estimate how much the extra bit is; guide them to report, “It’s a little more than a tenth of the bicycle diameter.”

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7•3 Lesson 16


The circumference of any circle is a little more than 3 times its diameter. The number pi is a little greater than 3.

Use the symbol 𝜋𝜋 to represent this special number. Pi is a non-terminating, non-repeating decimal, and mathematicians use the symbol 𝜋𝜋 or approximate representations as more convenient ways to represent pi.

Mathematical Modeling Exercise

The ratio of the circumference to its diameter is always the same for any circle. The value of this ratio, 𝐂𝐂𝐂𝐂𝐂𝐂𝐜𝐜𝐂𝐂𝐜𝐜𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐜𝐜𝐂𝐂

𝐃𝐃𝐂𝐂𝐃𝐃𝐜𝐜𝐂𝐂𝐃𝐃𝐂𝐂𝐂𝐂, is

called the number pi and is represented by the symbol 𝝅𝝅.

Since the circumference is a little greater than 𝟑𝟑 times the diameter, 𝝅𝝅 is a number that is a little greater than 𝟑𝟑. Use the symbol 𝝅𝝅 to represent this special number. Pi is a non-terminating, non-repeating decimal, and mathematicians use the symbol 𝝅𝝅 or approximate representations as more convenient ways to represent pi.

• 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟒𝟒 or 𝟐𝟐𝟐𝟐𝟕𝟕

.

• The ratios of the circumference to the diameter and 𝝅𝝅 ∶ 𝟏𝟏 are equal.

• 𝐂𝐂𝐂𝐂𝐂𝐂𝐜𝐜𝐂𝐂𝐜𝐜𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐜𝐜𝐂𝐂 𝐨𝐨𝐂𝐂 𝐃𝐃 𝐂𝐂𝐂𝐂𝐂𝐂𝐜𝐜𝐂𝐂𝐂𝐂 = 𝝅𝝅 × 𝐃𝐃𝐂𝐂𝐃𝐃𝐜𝐜𝐂𝐂𝐃𝐃𝐂𝐂𝐂𝐂.

Example (10 minutes)

Note that both 3.14 and 227 are excellent approximations to use in the classroom: One helps students’ fluency with

decimal number arithmetic, and the second helps students’ fluency with fraction arithmetic. After learning about 𝜋𝜋 and its approximations, have students use the 𝜋𝜋 button on their calculators as another approximation for 𝜋𝜋. Students should use all digits of 𝜋𝜋 in the calculator and round appropriately.

Example

a. The following circles are not drawn to scale. Find the circumference of each circle. (Use 𝟐𝟐𝟐𝟐𝟕𝟕 as an approximation for 𝝅𝝅.)

𝟔𝟔𝟔𝟔 𝐜𝐜𝐜𝐜; 𝟐𝟐𝟐𝟐𝟔𝟔 𝐂𝐂𝐃𝐃.; 𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜; Ask students if these numbers are roughly three times the diameters.

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7•3 Lesson 16


b. The radius of a paper plate is 𝟏𝟏𝟏𝟏.𝟕𝟕 𝐜𝐜𝐜𝐜. Find the circumference to the nearest tenth. (Use 𝟑𝟑.𝟏𝟏𝟒𝟒 as an approximation for 𝝅𝝅.)

Diameter: 𝟐𝟐𝟑𝟑.𝟒𝟒 𝐜𝐜𝐜𝐜; circumference: 𝟕𝟕𝟑𝟑.𝟓𝟓 𝐜𝐜𝐜𝐜

Extension for this problem: Bring in paper plates, and ask students how to find the center of a paper plate. This is not as easy as it sounds because the center is not given. Answer: Fold the paper plate in half twice. The intersection of the two folds is the center. Afterward, have students fold their paper plates several more times. Explore what happens. Ask students why the intersection of both lines is guaranteed to be the center. Answer: The first fold guarantees that the crease is a diameter, the second fold divides that diameter in half, but the midpoint of a diameter is the center.

c. The radius of a paper plate is 𝟏𝟏𝟏𝟏.𝟕𝟕 𝐜𝐜𝐜𝐜. Find the circumference to the nearest hundredth. (Use the 𝝅𝝅 button on your calculator as an approximation for 𝝅𝝅.)

Circumference: 𝟕𝟕𝟑𝟑.𝟓𝟓𝟏𝟏 𝐜𝐜𝐜𝐜

d. A circle has a radius of 𝒓𝒓 𝐜𝐜𝐜𝐜 and a circumference of 𝑪𝑪 𝐜𝐜𝐜𝐜. Write a formula that expresses the value of 𝑪𝑪 in terms of 𝒓𝒓 and 𝝅𝝅.

𝑪𝑪 = 𝝅𝝅 ∙ 𝟐𝟐𝒓𝒓, or 𝑪𝑪 = 𝟐𝟐𝝅𝝅𝒓𝒓.

e. The figure below is in the shape of a semicircle. A semicircle is an arc that is half of a circle. Find the perimeter of the shape. (Use 𝟑𝟑.𝟏𝟏𝟒𝟒 for 𝝅𝝅.)

𝟐𝟐 𝐜𝐜 +𝟐𝟐(𝟑𝟑.𝟏𝟏𝟒𝟒)

𝟐𝟐 𝐜𝐜 = 𝟐𝟐𝟏𝟏.𝟓𝟓𝟔𝟔 𝐜𝐜

Closing (5 minutes) Relevant Vocabulary

CIRCLE: Given a point 𝑶𝑶 in the plane and a number 𝒓𝒓 > 𝟏𝟏, the circle with center 𝑶𝑶 and radius 𝒓𝒓 is the set of all points in the plane whose distance from the point 𝑶𝑶 is equal to 𝒓𝒓.

RADIUS OF A CIRCLE: The radius is the length of any segment whose endpoints are the center of a circle and a point that lies on the circle.

DIAMETER OF A CIRCLE: The diameter of a circle is the length of any segment that passes through the center of a circle whose endpoints lie on the circle. If 𝒓𝒓 is the radius of a circle, then the diameter is 𝟐𝟐𝒓𝒓.

The word diameter can also mean the segment itself. Context determines how the term is being used: The diameter usually refers to the length of the segment, while a diameter usually refers to a segment. Similarly, a radius can refer to a segment from the center of a circle to a point on the circle.

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Circumference

CIRCUMFERENCE: The circumference of a circle is the distance around a circle.

PI: The number pi, denoted by 𝝅𝝅, is the value of the ratio given by the circumference to the diameter, that is

𝝅𝝅 = 𝐜𝐜𝐂𝐂𝐂𝐂𝐜𝐜𝐂𝐂𝐜𝐜𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐜𝐜𝐂𝐂𝐝𝐝𝐂𝐂𝐃𝐃𝐜𝐜𝐂𝐂𝐃𝐃𝐂𝐂𝐂𝐂 . The most commonly used approximations for 𝝅𝝅 is 𝟑𝟑.𝟏𝟏𝟒𝟒 or 𝟐𝟐𝟐𝟐𝟕𝟕 .

SEMICIRCLE: Let 𝑪𝑪 be a circle with center 𝑶𝑶, and let 𝑨𝑨 and 𝑩𝑩 be the endpoints of a diameter. A semicircle is the set containing 𝑨𝑨, 𝑩𝑩, and all points that lie in a given half-plane determined by 𝑨𝑨𝑩𝑩�� (diameter) that lie on circle 𝑪𝑪.


The Exit Ticket calls on students to synthesize their knowledge of circles and rectangles. A simpler alternative is to have students sketch a circle with a given radius and then have them determine the diameter and circumference of that circle.

Semicircle

Circle C

Radii: 𝑶𝑶𝑨𝑨��,𝑶𝑶𝑩𝑩��,𝑶𝑶𝑶𝑶��

Diameter: 𝑨𝑨𝑩𝑩��

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7•3 Lesson 16


Name ___________________________________________________ Date____________________


Exit Ticket Brianna’s parents built a swimming pool in the backyard. Brianna says that the distance around the pool is 120 feet.

1. Is she correct? Explain why or why not.

2. Explain how Brianna would determine the distance around the pool so that her parents would know how many feet of stone to buy for the edging around the pool.

3. Explain the relationship between the circumference of the semicircular part of the pool and the width of the pool.

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Brianna’s parents built a swimming pool in the backyard. Brianna says that the distance around the pool is 𝟏𝟏𝟐𝟐𝟏𝟏 feet.

1. Is she correct? Explain why or why not.

Brianna is incorrect. The distance around the pool is 𝟏𝟏𝟑𝟑𝟏𝟏.𝟒𝟒 𝐂𝐂𝐃𝐃. She found the distance around the rectangle only and did not include the distance around the semicircular part of the pool.

2. Explain how Brianna would determine the distance around the pool so that her parents would know how many feet of stone to buy for the edging around the pool.

In order to find the distance around the pool, Brianna must first find the circumference of the semicircle, which is

𝑪𝑪 = 𝟏𝟏𝟐𝟐 ∙ 𝝅𝝅 ∙ 𝟐𝟐𝟏𝟏 𝐂𝐂𝐃𝐃., or 𝟏𝟏𝟏𝟏𝝅𝝅 𝐂𝐂𝐃𝐃., or about 𝟑𝟑𝟏𝟏.𝟒𝟒 𝐂𝐂𝐃𝐃. The sum of the three other sides is

𝟐𝟐𝟏𝟏 𝐂𝐂𝐃𝐃. + 𝟒𝟒𝟏𝟏 𝐂𝐂𝐃𝐃. + 𝟒𝟒𝟏𝟏 𝐂𝐂𝐃𝐃. = 𝟏𝟏𝟏𝟏𝟏𝟏 𝐂𝐂𝐃𝐃.; the perimeter is 𝟏𝟏𝟏𝟏𝟏𝟏 𝐂𝐂𝐃𝐃. + 𝟑𝟑𝟏𝟏.𝟒𝟒 𝐂𝐂𝐃𝐃. = 𝟏𝟏𝟑𝟑𝟏𝟏.𝟒𝟒 𝐂𝐂𝐃𝐃.

3. Explain the relationship between the circumference of the semicircular part of the pool and the width of the pool.

The relationship between the circumference of the semicircular part and the width of the pool is the same as half of 𝝅𝝅 because this is half the circumference of the entire circle.

Problem Set Sample Solutions Students should work in cooperative groups to complete the tasks for this exercise.

1. Find the circumference.

a. Give an exact answer in terms of 𝝅𝝅.

𝑪𝑪 = 𝟐𝟐𝝅𝝅𝒓𝒓 𝑪𝑪 = 𝟐𝟐𝝅𝝅 ∙ 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜 𝑪𝑪 = 𝟐𝟐𝟐𝟐𝝅𝝅 𝐜𝐜𝐜𝐜

b. Use 𝝅𝝅 ≈ 𝟐𝟐𝟐𝟐𝟕𝟕 , and express your answer as a fraction in lowest terms.

𝑪𝑪 ≈ 𝟐𝟐 ∙𝟐𝟐𝟐𝟐𝟕𝟕∙ 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜

𝑪𝑪 ≈ 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜

c. Use 𝒕𝒕𝒕𝒕𝒕𝒕 𝝅𝝅 button on your calculator, and express your answer to the nearest hundredth.

𝑪𝑪 = 𝟐𝟐 ∙ 𝝅𝝅 ∙ 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜

𝑪𝑪 ≈ 𝟐𝟐𝟕𝟕.𝟗𝟗𝟔𝟔 𝐜𝐜𝐜𝐜

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2. Find the circumference.

a. Give an exact answer in terms of 𝝅𝝅.

𝒅𝒅 = 𝟒𝟒𝟐𝟐 𝐜𝐜𝐜𝐜

𝑪𝑪 = 𝝅𝝅𝒅𝒅 𝑪𝑪 = 𝟒𝟒𝟐𝟐𝝅𝝅 𝐜𝐜𝐜𝐜

b. Use 𝝅𝝅 ≈ 𝟐𝟐𝟐𝟐𝟕𝟕 , and express your answer as a fraction in lowest terms.

𝑪𝑪 ≈ 𝟒𝟒𝟐𝟐 𝐜𝐜𝐜𝐜 ∙𝟐𝟐𝟐𝟐𝟕𝟕

𝑪𝑪 ≈ 𝟏𝟏𝟑𝟑𝟐𝟐 𝐜𝐜𝐜𝐜

3. The figure shows a circle within a square. Find the circumference of the circle. Let 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟒𝟒.

4. Consider the diagram of a semicircle shown.

a. Explain in words how to determine the perimeter of a semicircle.

The perimeter is the sum of the length of the diameter and half of the circumference of a circle with the same diameter.

b. Using 𝒅𝒅 to represent the diameter of the circle, write an algebraic equation that will result in the perimeter of a semicircle.

𝑷𝑷 = 𝒅𝒅 + 𝟏𝟏𝟐𝟐𝝅𝝅𝒅𝒅

c. Write another algebraic equation to represent the perimeter of a semicircle using 𝒓𝒓 to represent the radius of a semicircle.

𝑷𝑷 = 𝟐𝟐𝒓𝒓 +𝟏𝟏𝟐𝟐𝝅𝝅 ∙ 𝟐𝟐𝒓𝒓

𝑷𝑷 = 𝟐𝟐𝒓𝒓 + 𝝅𝝅𝒓𝒓

𝟏𝟏𝟔𝟔 𝐂𝐂𝐂𝐂.

The diameter of the circle is the same as the length of the side of the square.

𝑪𝑪 = 𝝅𝝅𝒅𝒅 𝑪𝑪 = 𝝅𝝅 ∙ 𝟏𝟏𝟔𝟔 𝐂𝐂𝐂𝐂. 𝑪𝑪 ≈ 𝟑𝟑.𝟏𝟏𝟒𝟒 ∙ 𝟏𝟏𝟔𝟔 𝐂𝐂𝐂𝐂. 𝑪𝑪 ≈ 𝟓𝟓𝟏𝟏.𝟐𝟐𝟒𝟒 𝐂𝐂𝐂𝐂.

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5. Find the perimeter of the semicircle. Let 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟒𝟒.

6. Ken’s landscape gardening business makes odd-shaped lawns that include semicircles. Find the length of the edging material needed to border the two lawn designs. Use 𝟑𝟑.𝟏𝟏𝟒𝟒 for 𝝅𝝅.

a. The radius of this flowerbed is 𝟐𝟐.𝟓𝟓 𝐜𝐜.

A semicircle has half of the circumference of a circle. If the circumference of the semicircle is

𝑪𝑪 = 𝟏𝟏𝟐𝟐 (𝝅𝝅 ∙ 𝟐𝟐 ∙ 𝟐𝟐.𝟓𝟓 𝐜𝐜), then the circumference approximates 𝟕𝟕.𝟐𝟐𝟓𝟓 𝐜𝐜. The length of the edging material must

include the circumference and the diameter; 𝟕𝟕.𝟐𝟐𝟓𝟓 𝐜𝐜+ 𝟓𝟓 𝐜𝐜 = 𝟏𝟏𝟐𝟐.𝟐𝟐𝟓𝟓 𝐜𝐜. Ken needs 𝟏𝟏𝟐𝟐.𝟐𝟐𝟓𝟓 meters of edging to complete his design.

b. The diameter of the semicircular section is 𝟏𝟏𝟏𝟏 𝐜𝐜, and the lengths of the two sides are 𝟔𝟔 𝐜𝐜.

The circumference of the semicircular part has half of the circumference of a circle. The circumference of the

semicircle is 𝐂𝐂 = 𝟏𝟏𝟐𝟐𝛑𝛑 ∙ 𝟏𝟏𝟏𝟏 𝐜𝐜, which is approximately 𝟏𝟏𝟓𝟓.𝟕𝟕 𝐜𝐜. The length of the edging material must include

the circumference of the semicircle and the perimeter of two sides of the triangle; 𝟏𝟏𝟓𝟓.𝟕𝟕 𝐜𝐜 + 𝟔𝟔 𝐜𝐜+ 𝟔𝟔 𝐜𝐜 = 𝟐𝟐𝟕𝟕.𝟕𝟕 𝐜𝐜. Ken needs 𝟐𝟐𝟕𝟕.𝟕𝟕 meters of edging to complete his design.

7. Mary and Margaret are looking at a map of a running path in a local park. Which is the shorter path from 𝑬𝑬 to 𝑪𝑪, along the two semicircles or along the larger semicircle? If one path is shorter, how much shorter is it? Let 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟒𝟒.

A semicircle has half of the circumference of a circle. The circumference of the large semicircle is 𝑪𝑪 = 𝟏𝟏𝟐𝟐𝝅𝝅 ∙ 𝟒𝟒 𝐤𝐤𝐜𝐜, or

𝟔𝟔.𝟐𝟐𝟐𝟐 𝐤𝐤𝐜𝐜. The diameter of the two smaller semicircles is 𝟐𝟐 𝐤𝐤𝐜𝐜. The total circumference would be the same as the circumference for a whole circle with the same diameter. If 𝑪𝑪 = 𝝅𝝅 ∙ 𝟐𝟐 𝐤𝐤𝐜𝐜, then 𝑪𝑪 = 𝟔𝟔.𝟐𝟐𝟐𝟐 𝐤𝐤𝐜𝐜. The distance around the larger semicircle is the same as the distance around both of the semicircles. So, both paths are equal in distance.

𝑷𝑷 = 𝒅𝒅+𝟏𝟏𝟐𝟐𝝅𝝅𝒅𝒅

𝑷𝑷 ≈ 𝟏𝟏𝟕𝟕 𝐂𝐂𝐂𝐂. +𝟏𝟏𝟐𝟐∙ 𝟑𝟑.𝟏𝟏𝟒𝟒 ∙ 𝟏𝟏𝟕𝟕 𝐂𝐂𝐂𝐂.

𝑷𝑷 ≈ 𝟏𝟏𝟕𝟕 𝐂𝐂𝐂𝐂. + 𝟐𝟐𝟔𝟔.𝟔𝟔𝟗𝟗 𝐂𝐂𝐂𝐂. 𝑷𝑷 ≈ 𝟒𝟒𝟑𝟑.𝟔𝟔𝟗𝟗 𝐂𝐂𝐂𝐂.

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8. Alex the electrician needs 𝟑𝟑𝟒𝟒 yards of electrical wire to complete a job. He has a coil of wiring in his workshop. The coiled wire is 𝟏𝟏𝟐𝟐 inches in diameter and is made up of 𝟐𝟐𝟏𝟏 circles of wire. Will this coil be enough to complete the job? Let 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟒𝟒.

The circumference of the coil of wire is 𝑪𝑪 = 𝝅𝝅 ∙ 𝟏𝟏𝟐𝟐 𝐂𝐂𝐂𝐂., or approximately 𝟓𝟓𝟔𝟔.𝟓𝟓𝟐𝟐 𝐂𝐂𝐂𝐂. If there are 𝟐𝟐𝟏𝟏 circles of wire, then the number of circles times the circumference will yield the total number of inches of wire in the coil. If

𝟓𝟓𝟔𝟔.𝟓𝟓𝟐𝟐 𝐂𝐂𝐂𝐂. ∙ 𝟐𝟐𝟏𝟏 ≈ 𝟏𝟏𝟏𝟏𝟐𝟐𝟔𝟔.𝟗𝟗𝟐𝟐 𝐂𝐂𝐂𝐂., then 𝟏𝟏𝟏𝟏𝟐𝟐𝟔𝟔.𝟗𝟗𝟐𝟐 𝐂𝐂𝐂𝐂.

𝟑𝟑𝟔𝟔 𝐂𝐂𝐂𝐂.≈ 𝟑𝟑𝟐𝟐. 𝟗𝟗𝟕𝟕 𝐲𝐲𝐝𝐝. (𝟏𝟏 𝐲𝐲𝐝𝐝. = 𝟑𝟑 𝐂𝐂𝐃𝐃. = 𝟑𝟑𝟔𝟔 𝐂𝐂𝐂𝐂. When converting inches to

yards, you must divide the total inches by the number of inches in a yard, which is 𝟑𝟑𝟔𝟔 inches.) Alex will not have enough wire for his job in this coil of wire.

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7•3 Lesson 17

Lesson 17: The Area of a Circle


Student Outcomes

Students give an informal derivation of the relationship between the circumference and area of a circle. Students know the formula for the area of a circle and use it to solve problems.

Lesson Notes Remind students of the definitions for circle and circumference from the previous lesson. The Opening

Exercise is a lead-in to the derivation of the formula for the area of a circle. Not only do students need to know and be able to apply the formula for the area of a circle, it is critical for

them to also be able to draw the diagram associated with each problem in order to solve it successfully.

Students must be able to translate words into mathematical expressions and equations and be able to determine which parts of the problem are known and which are unknown or missing.

Classwork

Exercises 1–3 (4 minutes) Exercises 1–3

Solve the problem below individually. Explain your solution.

1. Find the radius of a circle if its circumference is 𝟑𝟑𝟑𝟑.𝟔𝟔𝟔𝟔 inches. Use 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏.

If 𝑪𝑪 = 𝟐𝟐𝝅𝝅𝟐𝟐, then 𝟑𝟑𝟑𝟑.𝟔𝟔𝟔𝟔 = 𝟐𝟐𝝅𝝅𝟐𝟐. Solving the equation for 𝟐𝟐,

𝟑𝟑𝟑𝟑.𝟔𝟔𝟔𝟔 = 𝟐𝟐𝝅𝝅𝟐𝟐

�𝟏𝟏𝟐𝟐𝝅𝝅

�𝟑𝟑𝟑𝟑.𝟔𝟔𝟔𝟔 = �𝟏𝟏𝟐𝟐𝝅𝝅

�𝟐𝟐𝝅𝝅𝟐𝟐

𝟏𝟏

𝟔𝟔.𝟐𝟐𝟔𝟔(𝟑𝟑𝟑𝟑.𝟔𝟔𝟔𝟔) ≈ 𝟐𝟐

𝟔𝟔 ≈ 𝟐𝟐

The radius of the circle is approximately 𝟔𝟔 𝐢𝐢𝐢𝐢.

2. Determine the area of the rectangle below. Name two ways that can be used to find the area of the rectangle.

The area of the rectangle is 𝟐𝟐𝟏𝟏 𝐜𝐜𝐦𝐦𝟐𝟐. The area can be found by counting the square units inside the rectangle or by multiplying the length (𝟔𝟔 𝐜𝐜𝐦𝐦) by the width (𝟏𝟏 𝐜𝐜𝐦𝐦).

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3. Find the length of a rectangle if the area is 𝟐𝟐𝟑𝟑 𝐜𝐜𝐦𝐦𝟐𝟐 and the width is 𝟑𝟑 𝐜𝐜𝐦𝐦.

If the area of the rectangle is 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝐥𝐥𝐀𝐀𝐢𝐢𝐥𝐥𝐥𝐥𝐥𝐥 ∙ 𝐰𝐰𝐢𝐢𝐰𝐰𝐥𝐥𝐥𝐥, then

𝟐𝟐𝟑𝟑 𝐜𝐜𝐦𝐦𝟐𝟐 = 𝐥𝐥 ∙ 𝟑𝟑 𝐜𝐜𝐦𝐦 𝟏𝟏𝟑𝟑⋅ 𝟐𝟐𝟑𝟑 𝐜𝐜𝐦𝐦𝟐𝟐 =

𝟏𝟏𝟑𝟑⋅ 𝒍𝒍 ⋅ 𝟑𝟑 𝐜𝐜𝐦𝐦

𝟗𝟗 𝐜𝐜𝐦𝐦 = 𝒍𝒍

Exploratory Challenge (10 minutes)

Complete the exercise below.

Exploratory Challenge

To find the formula for the area of a circle, cut a circle into 𝟏𝟏𝟔𝟔 equal pieces.

Arrange the triangular wedges by alternating the “triangle” directions and sliding them together to make a “parallelogram.” Cut the triangle on the left side in half on the given line, and slide the outside half of the triangle to the other end of the parallelogram in order to create an approximate “rectangle.”

The circumference is 𝟐𝟐𝝅𝝅𝟐𝟐, where the radius is 𝟐𝟐. Therefore, half of the circumference is 𝝅𝝅𝟐𝟐.

Move half to the other end.

Scaffolding: Provide a circle divided into 16 equal sections for students to cut out and reassemble as a rectangle.

MP.7

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7•3 Lesson 17


What is the area of the “rectangle” using the side lengths above?

The area of the “rectangle” is base times height, and, in this case, 𝑨𝑨 = 𝝅𝝅𝟐𝟐 ∙ 𝟐𝟐.

Are the areas of the “rectangle” and the circle the same?

Yes, since we just rearranged pieces of the circle to make the “rectangle,” the area of the “rectangle” and the area of the circle are approximately equal. Note that the more sections we cut the circle into, the closer the approximation.

If the area of the rectangular shape and the circle are the same, what is the area of the circle?

The area of a circle is written as 𝑨𝑨 = 𝝅𝝅𝟐𝟐 ∙ 𝟐𝟐, or 𝑨𝑨 = 𝝅𝝅𝟐𝟐𝟐𝟐.


Example 1

Use the shaded square centimeter units to approximate the area of the circle.

What is the radius of the circle?

𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦

What would be a quicker method for determining the area of the circle other than counting all of the squares in the entire circle?

Count 𝟏𝟏𝟏𝟏

of the squares needed; then, multiply that by four in order to determine the area of the entire circle.

Using the diagram, how many squares were used to cover one-fourth of the circle?

The area of one-fourth of the circle is approximately 𝟑𝟑𝟗𝟗 𝐜𝐜𝐦𝐦𝟐𝟐.

What is the area of the entire circle?

𝑨𝑨 ≈ 𝟏𝟏 ∙ 𝟑𝟑𝟗𝟗 𝐜𝐜𝐦𝐦𝟐𝟐 𝑨𝑨 ≈ 𝟑𝟑𝟏𝟏𝟔𝟔 𝐜𝐜𝐦𝐦𝟐𝟐

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Example 2

A sprinkler rotates in a circular pattern and sprays water over a distance of 𝟏𝟏𝟐𝟐 feet. What is the area of the circular region covered by the sprinkler? Express your answer to the nearest square foot.

Draw a diagram to assist you in solving the problem. What does the distance of 𝟏𝟏𝟐𝟐 feet represent in this problem?

The radius is 𝟏𝟏𝟐𝟐 feet.

What information is needed to solve the problem?

The formula to find the area of a circle is 𝑨𝑨 = 𝝅𝝅𝟐𝟐𝟐𝟐. If the radius is 𝟏𝟏𝟐𝟐 𝐟𝐟𝐥𝐥., then 𝑨𝑨 = 𝝅𝝅 ∙ (𝟏𝟏𝟐𝟐 𝐟𝐟𝐥𝐥. )𝟐𝟐 = 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅 𝐟𝐟𝐥𝐥𝟐𝟐, or approximately 𝟏𝟏𝟒𝟒𝟐𝟐 𝐟𝐟𝐥𝐥𝟐𝟐.

Make a point of telling students that an answer in exact form is in terms of 𝜋𝜋, not substituting an approximation of pi.


Example 3

Suzanne is making a circular table out of a square piece of wood. The radius of the circle that she is cutting is 𝟑𝟑 feet. How much waste will she have for this project? Express your answer to the nearest square foot.

Draw a diagram to assist you in solving the problem. What does the distance of 𝟑𝟑 feet represent in this problem?

The radius of the circle is 𝟑𝟑 feet.


The area of the circle and the area of the square are needed so that we can subtract the area of the circle from the area of the square to determine the amount of waste.

What information do we need to determine the area of the square and the circle?

Circle: just radius because 𝑨𝑨 = 𝝅𝝅𝟐𝟐𝟐𝟐 Square: one side length

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7•3 Lesson 17


How will we determine the waste?

The waste is the area left over from the square after cutting out the circular region. The area of the circle is 𝑨𝑨 = 𝝅𝝅 ∙ (𝟑𝟑 𝐟𝐟𝐥𝐥. )𝟐𝟐 = 𝟗𝟗𝝅𝝅 𝐟𝐟𝐥𝐥𝟐𝟐 ≈ 𝟐𝟐𝟔𝟔.𝟐𝟐𝟔𝟔 𝐟𝐟𝐥𝐥𝟐𝟐. The area of the square is found by first finding the diameter of the circle, which is the same as the side of the square. The diameter is 𝒅𝒅 = 𝟐𝟐𝟐𝟐; so, 𝒅𝒅 = 𝟐𝟐 ∙ 𝟑𝟑 𝐟𝐟𝐥𝐥. or 𝟔𝟔 𝐟𝐟𝐥𝐥. The area of a square is found by multiplying the length and width, so 𝑨𝑨 = 𝟔𝟔 𝐟𝐟𝐥𝐥. ∙ 𝟔𝟔 𝐟𝐟𝐥𝐥. = 𝟑𝟑𝟔𝟔 𝐟𝐟𝐥𝐥𝟐𝟐. The solution is the difference between the area of the square and the area of the circle, so 𝟑𝟑𝟔𝟔 𝐟𝐟𝐥𝐥𝟐𝟐 − 𝟐𝟐𝟔𝟔.𝟐𝟐𝟔𝟔 𝐟𝐟𝐥𝐥𝟐𝟐 ≈ 𝟑𝟑.𝟑𝟑𝟏𝟏 𝐟𝐟𝐥𝐥𝟐𝟐.

Does your solution answer the problem as stated?

Yes, the amount of waste is 𝟑𝟑.𝟑𝟑𝟏𝟏 𝐟𝐟𝐥𝐥𝟐𝟐.

Exercises 4–6 (11 minutes)

Solve in cooperative groups of two or three.

Exercises 4–6

4. A circle has a radius of 𝟐𝟐 cm.

a. Find the exact area of the circular region.

𝐀𝐀 = 𝛑𝛑 ∙ (𝟐𝟐 𝐜𝐜𝐦𝐦)𝟐𝟐 = 𝟏𝟏𝛑𝛑 𝐜𝐜𝐦𝐦𝟐𝟐

b. Find the approximate area using 𝟑𝟑.𝟏𝟏𝟏𝟏 to approximate 𝛑𝛑.

𝐀𝐀 = 𝟏𝟏 𝐜𝐜𝐦𝐦𝟐𝟐 ∙ 𝛑𝛑 ≈ 𝟏𝟏 𝐜𝐜𝐦𝐦𝟐𝟐 ∙ 𝟑𝟑.𝟏𝟏𝟏𝟏 ≈ 𝟏𝟏𝟐𝟐.𝟒𝟒𝟔𝟔 𝐜𝐜𝐦𝐦𝟐𝟐

5. A circle has a radius of 𝟑𝟑 cm.

a. Find the exact area of the circular region.

𝐀𝐀 = 𝛑𝛑 ∙ (𝟑𝟑 𝐜𝐜𝐦𝐦)𝟐𝟐 = 𝟏𝟏𝟗𝟗𝛑𝛑 𝐜𝐜𝐦𝐦𝟐𝟐

b. Find the approximate area using 𝟐𝟐𝟐𝟐𝟑𝟑

to approximate 𝛑𝛑.

𝐀𝐀 = 𝟏𝟏𝟗𝟗 ∙ 𝛑𝛑 𝐜𝐜𝐦𝐦𝟐𝟐 ≈ �𝟏𝟏𝟗𝟗 ∙𝟐𝟐𝟐𝟐𝟑𝟑� 𝐜𝐜𝐦𝐦𝟐𝟐 ≈ 𝟏𝟏𝟒𝟒𝟏𝟏 𝐜𝐜𝐦𝐦𝟐𝟐

c. What is the circumference of the circle?

𝐂𝐂 = 𝟐𝟐𝛑𝛑 ∙ 𝟑𝟑 𝐜𝐜𝐦𝐦 = 𝟏𝟏𝟏𝟏𝛑𝛑 𝐜𝐜𝐦𝐦 ≈ 𝟏𝟏𝟑𝟑.𝟗𝟗𝟔𝟔 𝐜𝐜𝐦𝐦

6. Joan determined that the area of the circle below is 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐. Melinda says that Joan’s solution is incorrect; she believes that the area is 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐. Who is correct and why?

Melinda is correct. Joan found the area by multiplying 𝝅𝝅 by the square of 𝟐𝟐𝟏𝟏 𝐜𝐜𝐦𝐦 (which is the diameter) to get a result of 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐, which is incorrect. Melinda found that the radius was 𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦 (half of the diameter). Melinda multiplied 𝝅𝝅 by the square of the radius to get a result of 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐.

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Closing (4 minutes)

Strategies for problem solving include drawing a diagram to represent the problem and identifying the given information and needed information to solve the problem.

Using the original circle in this lesson, cut it into 64 equal slices. Reassemble the figure. What do you notice?

It looks more like a rectangle.

Ask students to imagine repeating the slicing into even thinner slices (infinitely thin). Then, ask the next two questions.

What does the length of the rectangle become?

An approximation of half of the circumference of the circle

What does the width of the rectangle become? An approximation of the radius

Thus, we conclude that the area of the circle is 𝐴𝐴 = 12𝐶𝐶𝐶𝐶.

• If 𝐴𝐴 = 12𝐶𝐶𝐶𝐶, then 𝐴𝐴 = 1

2 ∙ 2𝜋𝜋𝐶𝐶 ∙ 𝐶𝐶 or 𝐴𝐴 = 𝜋𝜋𝐶𝐶2.

• Also see video link: http://www.youtube.com/watch?v=YokKp3pwVFc

Relevant Vocabulary

CIRCULAR REGION (OR DISK): Given a point 𝑪𝑪 in the plane and a number 𝟐𝟐 > 𝟏𝟏, the circular region (or disk) with center 𝑪𝑪 and radius 𝟐𝟐 is the set of all points in the plane whose distance from the point 𝑪𝑪 is less than or equal to 𝟐𝟐.

The boundary of a disk is a circle. The area of a circle refers to the area of the disk defined by the circle.


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7•3 Lesson 17


Word bank

radius height base 𝟐𝟐𝝅𝝅𝟐𝟐

diameter circle rectangle 𝝅𝝅𝟐𝟐𝟐𝟐 𝝅𝝅

Name ___________________________________________________ Date____________________


Exit Ticket Complete each statement using the words or algebraic expressions listed in the word bank below.

1. The length of the of the rectangular region approximates the length of the of the circle.

2. The of the rectangle approximates the length of one-half of the circumference of the circle.

3. The circumference of the circle is _______________________.

4. The _________________ of the ___________________ is 2𝐶𝐶.

5. The ratio of the circumference to the diameter is ______.

6. Area (circle) = Area of (_____________) = 12 ∙ circumference ∙ 𝐶𝐶 = 1

2 (2𝜋𝜋𝐶𝐶) ∙ 𝐶𝐶 = 𝜋𝜋 ∙ 𝐶𝐶 ∙ 𝐶𝐶 = _____________.

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Exit Ticket Sample Solutions Complete each statement using the words or algebraic expressions listed in the word bank below.

1. The length of the height of the rectangular region approximates the length of the radius of the circle.

2. The base of the rectangle approximates the length of one-half of the circumference of the circle.

3. The circumference of the circle is 𝟐𝟐𝛑𝛑𝐀𝐀.

4. The diameter of the circle is 𝟐𝟐𝐀𝐀.

5. The ratio of the circumference to the diameter is 𝝅𝝅.

6. Area (circle) = Area of (rectangle) = 𝟏𝟏𝟐𝟐 ∙ 𝐜𝐜𝐢𝐢𝐀𝐀𝐜𝐜𝐜𝐜𝐦𝐦𝐟𝐟𝐀𝐀𝐀𝐀𝐀𝐀𝐢𝐢𝐜𝐜𝐀𝐀 ∙ 𝟐𝟐 = 𝟏𝟏

𝟐𝟐 (𝟐𝟐𝝅𝝅𝟐𝟐) ∙ 𝟐𝟐 = 𝝅𝝅 ∙ 𝟐𝟐 ∙ 𝟐𝟐 = 𝛑𝛑𝐀𝐀𝟐𝟐.


1. The following circles are not drawn to scale. Find the area of each circle. (Use 𝟐𝟐𝟐𝟐𝟑𝟑

as an approximation for 𝝅𝝅.)

𝟑𝟑𝟏𝟏𝟔𝟔.𝟒𝟒 𝐜𝐜𝐦𝐦𝟐𝟐 𝟒𝟒,𝟏𝟏𝟒𝟒𝟒𝟒.𝟏𝟏 𝐟𝐟𝐥𝐥𝟐𝟐 𝟏𝟏,𝟒𝟒𝟗𝟗𝟏𝟏.𝟏𝟏 𝐜𝐜𝐦𝐦𝟐𝟐

2. A circle has a diameter of 𝟐𝟐𝟏𝟏 inches.

a. Find the exact area, and find an approximate area using 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏.

If the diameter is 𝟐𝟐𝟏𝟏 𝐢𝐢𝐢𝐢., then the radius is 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. If 𝐀𝐀 = 𝛑𝛑𝐀𝐀𝟐𝟐, then 𝐀𝐀 = 𝛑𝛑 ∙ (𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. )𝟐𝟐 or 𝟏𝟏𝟏𝟏𝟏𝟏𝛑𝛑 𝐢𝐢𝐢𝐢𝟐𝟐. 𝐀𝐀 ≈ (𝟏𝟏𝟏𝟏𝟏𝟏 ∙ 𝟑𝟑.𝟏𝟏𝟏𝟏) 𝐢𝐢𝐢𝐢𝟐𝟐 ≈ 𝟑𝟑𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐.

b. What is the circumference of the circle using 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏 ?

If the diameter is 𝟐𝟐𝟏𝟏 𝐢𝐢𝐢𝐢., then the circumference is 𝐂𝐂 = 𝛑𝛑𝐰𝐰 or 𝐂𝐂 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏 ∙ 𝟐𝟐𝟏𝟏 𝐢𝐢𝐢𝐢. ≈ 𝟔𝟔𝟐𝟐.𝟔𝟔 𝐢𝐢𝐢𝐢.

3. A circle has a diameter of 𝟏𝟏𝟏𝟏 inches.

a. Find the exact area and an approximate area using 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏.

If the diameter is 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢., then the radius is 𝟏𝟏𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. If 𝐀𝐀 = 𝛑𝛑𝐀𝐀𝟐𝟐, then 𝐀𝐀 = 𝛑𝛑 ∙ �𝟏𝟏𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. �𝟐𝟐or 𝟏𝟏𝟐𝟐𝟏𝟏𝟏𝟏 𝛑𝛑 𝐢𝐢𝐢𝐢𝟐𝟐.

𝐀𝐀 ≈ �𝟏𝟏𝟐𝟐𝟏𝟏𝟏𝟏

∙ 𝟑𝟑.𝟏𝟏𝟏𝟏� 𝐢𝐢𝐢𝐢𝟐𝟐 ≈ 𝟗𝟗𝟏𝟏.𝟗𝟗𝟔𝟔𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐

b. What is the circumference of the circle using 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏?

If the diameter is 𝟏𝟏𝟏𝟏 inches, then the circumference is 𝐂𝐂 = 𝛑𝛑𝐰𝐰 or 𝐂𝐂 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏 ∙ 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢.≈ 𝟑𝟑𝟏𝟏.𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢.

𝟏𝟏𝟒𝟒𝟐𝟐𝐜𝐜𝐦𝐦

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4. Using the figure below, find the area of the circle.

In this circle, the diameter is the same as the length of the side of the square. The diameter is 𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦; so, the radius is 𝟒𝟒 𝐜𝐜𝐦𝐦. 𝑨𝑨 = 𝝅𝝅𝟐𝟐𝟐𝟐, so 𝑨𝑨 = 𝝅𝝅(𝟒𝟒 𝐜𝐜𝐦𝐦)𝟐𝟐 = 𝟐𝟐𝟒𝟒𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐.

5. A path bounds a circular lawn at a park. If the inner edge of the path is 𝟏𝟏𝟑𝟑𝟐𝟐 𝐟𝐟𝐥𝐥. around, approximate the amount of

area of the lawn inside the circular path. Use 𝝅𝝅 ≈ 𝟐𝟐𝟐𝟐𝟑𝟑 .

The length of the path is the same as the circumference. Find the radius from the circumference; then, find the area.

𝑪𝑪 = 𝟐𝟐𝝅𝝅𝟐𝟐

𝟏𝟏𝟑𝟑𝟐𝟐 𝐟𝐟𝐥𝐥.≈ 𝟐𝟐 ∙𝟐𝟐𝟐𝟐𝟑𝟑∙ 𝟐𝟐

𝟏𝟏𝟑𝟑𝟐𝟐 𝐟𝐟𝐥𝐥.≈𝟏𝟏𝟏𝟏𝟑𝟑𝟐𝟐

𝟑𝟑𝟏𝟏𝟏𝟏

∙ 𝟏𝟏𝟑𝟑𝟐𝟐 𝐟𝐟𝐥𝐥.≈𝟑𝟑𝟏𝟏𝟏𝟏

∙𝟏𝟏𝟏𝟏𝟑𝟑𝟐𝟐

𝟐𝟐𝟏𝟏 𝐟𝐟𝐥𝐥.≈ 𝟐𝟐

𝑨𝑨 ≈𝟐𝟐𝟐𝟐𝟑𝟑∙ (𝟐𝟐𝟏𝟏 𝐟𝐟𝐥𝐥. )𝟐𝟐

𝑨𝑨 ≈ 𝟏𝟏𝟑𝟑𝟔𝟔𝟔𝟔 𝐟𝐟𝐥𝐥𝟐𝟐

6. The area of a circle is 𝟑𝟑𝟔𝟔𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐. Find its circumference.

Find the radius from the area of the circle; then, use it to find the circumference.

𝑨𝑨 = 𝝅𝝅𝟐𝟐𝟐𝟐 𝟑𝟑𝟔𝟔𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐 = 𝝅𝝅𝟐𝟐𝟐𝟐

𝟏𝟏𝝅𝝅∙ 𝟑𝟑𝟔𝟔𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐 =

𝟏𝟏𝝅𝝅∙ 𝝅𝝅𝟐𝟐𝟐𝟐

𝟑𝟑𝟔𝟔 𝐜𝐜𝐦𝐦𝟐𝟐 = 𝟐𝟐𝟐𝟐 𝟔𝟔 𝐜𝐜𝐦𝐦 = 𝟐𝟐

𝑪𝑪 = 𝟐𝟐𝝅𝝅𝟐𝟐 𝑪𝑪 = 𝟐𝟐𝝅𝝅 ∙ 𝟔𝟔 𝐜𝐜𝐦𝐦 𝑪𝑪 = 𝟏𝟏𝟐𝟐𝝅𝝅 𝐜𝐜𝐦𝐦

7. Find the ratio of the area of two circles with radii 𝟑𝟑 𝐜𝐜𝐦𝐦 and 𝟏𝟏 𝐜𝐜𝐦𝐦.

The area of the circle with radius 𝟑𝟑 𝐜𝐜𝐦𝐦 is 𝟗𝟗𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐. The area of the circle with the radius 𝟏𝟏 𝐜𝐜𝐦𝐦 is 𝟏𝟏𝟔𝟔𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐. The ratio of the area of the two circles is 𝟗𝟗𝝅𝝅 ∶ 𝟏𝟏𝟔𝟔𝝅𝝅 or 𝟗𝟗 ∶ 𝟏𝟏𝟔𝟔.

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8. If one circle has a diameter of 𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦 and a second circle has a diameter of 𝟐𝟐𝟏𝟏 𝐜𝐜𝐦𝐦, what is the ratio of the area of the larger circle to the area of the smaller circle?

The area of the circle with the diameter of 𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦 has a radius of 𝟒𝟒 𝐜𝐜𝐦𝐦. The area of the circle with the diameter of 𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦 is 𝝅𝝅 ∙ (𝟒𝟒 𝐜𝐜𝐦𝐦)𝟐𝟐, or 𝟐𝟐𝟒𝟒𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐. The area of the circle with the diameter of 𝟐𝟐𝟏𝟏 𝐜𝐜𝐦𝐦 has a radius of 𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦. The area of the circle with the diameter of 𝟐𝟐𝟏𝟏 𝐜𝐜𝐦𝐦 is 𝝅𝝅 ∙ (𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦)𝟐𝟐 or 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐. The ratio of the diameters is 𝟐𝟐𝟏𝟏 to 𝟏𝟏𝟏𝟏 or 𝟐𝟐:𝟏𝟏, while the ratio of the areas is 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅 𝒕𝒕o 𝟐𝟐𝟒𝟒𝝅𝝅 or 𝟏𝟏:𝟏𝟏.

9. Describe a rectangle whose perimeter is 𝟏𝟏𝟑𝟑𝟐𝟐 𝐟𝐟𝐥𝐥. and whose area is less than 𝟏𝟏 𝐟𝐟𝐥𝐥𝟐𝟐. Is it possible to find a circle whose circumference is 𝟏𝟏𝟑𝟑𝟐𝟐 𝐟𝐟𝐥𝐥. and whose area is less than 𝟏𝟏 𝐟𝐟𝐥𝐥𝟐𝟐? If not, provide an example or write a sentence explaining why no such circle exists.

A rectangle that has a perimeter of 𝟏𝟏𝟑𝟑𝟐𝟐 𝐟𝐟𝐥𝐥. can have a length of 𝟔𝟔𝟒𝟒.𝟗𝟗𝟗𝟗𝟒𝟒 𝐟𝐟𝐥𝐥. and a width of 𝟏𝟏 .𝟏𝟏𝟏𝟏𝟒𝟒 𝐟𝐟𝐥𝐥. The area of such a rectangle is 𝟏𝟏.𝟑𝟑𝟐𝟐𝟗𝟗𝟗𝟗𝟑𝟑𝟒𝟒 𝐟𝐟𝐥𝐥𝟐𝟐, which is less than 𝟏𝟏 𝐟𝐟𝐥𝐥𝟐𝟐. No, because a circle that has a circumference of 𝟏𝟏𝟑𝟑𝟐𝟐 𝐟𝐟𝐥𝐥. has a radius of approximately 𝟐𝟐𝟏𝟏 𝐟𝐟𝐥𝐥.

𝑨𝑨 = 𝝅𝝅𝟐𝟐𝟐𝟐 = 𝝅𝝅(𝟐𝟐𝟏𝟏)𝟐𝟐 = 𝟏𝟏𝟑𝟑𝟔𝟔𝟑𝟑.𝟗𝟗𝟔𝟔 ≠ 𝟏𝟏

10. If the diameter of a circle is double the diameter of a second circle, what is the ratio of the area of the first circle to the area of the second?

If I choose a diameter of 𝟐𝟐𝟏𝟏 𝐜𝐜𝐦𝐦 for the first circle, then the diameter of the second circle is 𝟏𝟏𝟐𝟐 𝐜𝐜𝐦𝐦. The first circle has a radius of 𝟏𝟏𝟐𝟐 𝐜𝐜𝐦𝐦 and an area of 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐. The second circle has a radius of 𝟔𝟔 𝐜𝐜𝐦𝐦 and an area of 𝟑𝟑𝟔𝟔𝝅𝝅 𝐜𝐜𝐦𝐦𝟐𝟐. The ratio of the area of the first circle to the second is 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅 to 𝟑𝟑𝟔𝟔𝝅𝝅 , which is a 𝟏𝟏 to 𝟏𝟏 ratio. The ratio of the diameters is 𝟐𝟐, while the ratio of the areas is the square of 𝟐𝟐, or 𝟏𝟏.

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7•3 Lesson 18

Lesson 18: More Problems on Area and Circumference


Student Outcomes

Students examine the meaning of quarter circle and semicircle. Students solve area and perimeter problems for regions made out of rectangles, quarter circles, semicircles,

and circles, including solving for unknown lengths when the area or perimeter is given.

Classwork


Students use prior knowledge to find the area of circles, semicircles, and quarter circles and compare their areas to areas of squares and rectangles.

Opening Exercise

Draw a circle with a diameter of 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 and a square with a side length of 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 on grid paper. Determine the area of the square and the circle.

Area of square: 𝑨𝑨 = (𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜)𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏; Area of circle: 𝑨𝑨 = 𝝅𝝅 ∙ (𝟔𝟔 𝐜𝐜𝐜𝐜)𝟏𝟏 = 𝟑𝟑𝟔𝟔𝝅𝝅 𝐜𝐜𝐜𝐜𝟏𝟏

Brainstorm some methods for finding half the area of the square and half the area of the circle.

Some methods include folding in half and counting the grid squares and cutting each in half and counting the squares.

Find the area of half of the square and half of the circle, and explain to a partner how you arrived at the area.

The area of half of the square is 𝟕𝟕𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏. The area of half of the circle is 𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐜𝐜𝟏𝟏. Some students may count the squares; others may realize that half of the square is a rectangle with side lengths of 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 and 𝟔𝟔 𝐜𝐜𝐜𝐜 and use 𝑨𝑨 = 𝒍𝒍 ∙ 𝒘𝒘 to determine the area. Some students may fold the square vertically, and some may fold it horizontally. Some students will try to count the grid squares in the semicircle and find that it is easiest to take half of the area of the circle.

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What is the ratio of the new area to the original area for the square and for the circle?

The ratio of the areas of the rectangle (half of the square) to the square is 𝟕𝟕𝟏𝟏:𝟏𝟏𝟏𝟏𝟏𝟏 or 𝟏𝟏:𝟏𝟏. The ratio for the areas of the circles is 𝟏𝟏𝟏𝟏𝝅𝝅:𝟑𝟑𝟔𝟔𝝅𝝅 or 𝟏𝟏:𝟏𝟏.

Find the area of one-fourth of the square and one-fourth of the circle, first by folding and then by another method. What is the ratio of the new area to the original area for the square and for the circle?

Folding the square in half and then in half again will result in one-fourth of the original square. The resulting shape is a square with a side length of 𝟔𝟔 𝐜𝐜𝐜𝐜 and an area of 𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏. Repeating the same process for the circle will result in an area of 𝟗𝟗𝝅𝝅 𝐜𝐜𝐜𝐜𝟏𝟏. The ratio for the areas of the squares is 𝟑𝟑𝟔𝟔:𝟏𝟏𝟏𝟏𝟏𝟏 or 𝟏𝟏:𝟏𝟏. The ratio for the areas of the circles is 𝟗𝟗𝝅𝝅:𝟑𝟑𝟔𝟔𝝅𝝅 or 𝟏𝟏:𝟏𝟏.

Write an algebraic expression that expresses the area of a semicircle and the area of a quarter circle.

Semicircle: 𝑨𝑨 = 𝟏𝟏𝟏𝟏𝝅𝝅𝒓𝒓

𝟏𝟏; Quarter circle: 𝑨𝑨 = 𝟏𝟏𝟏𝟏𝝅𝝅𝒓𝒓

𝟏𝟏


Example 1

Find the area of the following semicircle. Use 𝝅𝝅 ≈ 𝟏𝟏𝟏𝟏𝟕𝟕 .

If the diameter of the circle is 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜, then the radius is 𝟕𝟕 𝐜𝐜𝐜𝐜. The area of the semicircle is half of the area of the circular region.

𝑨𝑨 ≈𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏𝟕𝟕∙ (𝟕𝟕 𝐜𝐜𝐜𝐜)𝟏𝟏

𝑨𝑨 ≈𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏𝟕𝟕∙ 𝟏𝟏𝟗𝟗 𝐜𝐜𝐜𝐜𝟏𝟏

𝑨𝑨 ≈ 𝟕𝟕𝟕𝟕 𝐜𝐜𝐜𝐜𝟏𝟏

What is the area of the quarter circle? Use 𝝅𝝅 ≈ 𝟏𝟏𝟏𝟏𝟕𝟕 .

𝑨𝑨 ≈𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏𝟕𝟕

(𝟔𝟔 𝐜𝐜𝐜𝐜)𝟏𝟏

𝑨𝑨 ≈𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏𝟕𝟕∙ 𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏

𝑨𝑨 ≈𝟏𝟏𝟗𝟗𝟏𝟏𝟕𝟕

𝐜𝐜𝐜𝐜𝟏𝟏

Let students reason out and vocalize that the area of a quarter circle must be one-fourth of the area of an entire circle.

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Discussion

Students should recognize that composition area problems involve the decomposition of the shapes that make up the entire region. It is also very important for students to understand that there are several perspectives in decomposing each shape and that there is not just one correct method. There is often more than one correct method; therefore, a student may feel that his solution (which looks different than the one other students present) is incorrect. Alleviate that anxiety by showing multiple correct solutions. For example, cut an irregular shape into squares and rectangles as seen below.


Example 2

Marjorie is designing a new set of placemats for her dining room table. She sketched a drawing of the placement on graph paper. The diagram represents the area of the placemat consisting of a rectangle and two semicircles at either end. Each square on the grid measures 𝟏𝟏 inches in length.

Find the area of the entire placemat. Explain your thinking regarding the solution to this problem.

The length of one side of the rectangular section is 𝟏𝟏𝟏𝟏 inches in length, while the width is 𝟏𝟏 inches. The radius of the semicircular region is 𝟏𝟏 inches. The area of the rectangular part is (𝟏𝟏 𝐢𝐢𝐢𝐢) ∙ (𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢) = 𝟗𝟗𝟔𝟔 𝐢𝐢𝐢𝐢𝟏𝟏. The total area must include the two semicircles on either end of the placemat. The area of the two semicircular regions is the same as the area of one circle with the same radius. The area of the circular region is 𝑨𝑨 = 𝝅𝝅 ∙ (𝟏𝟏 𝐢𝐢𝐢𝐢)𝟏𝟏 = 𝟏𝟏𝟔𝟔𝝅𝝅 𝐢𝐢𝐢𝐢𝟏𝟏. In this problem, using 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏 makes more sense because there are no fractions in the problem. The area of the semicircular regions is approximately 𝟓𝟓𝟓𝟓.𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏. The total area for the placemat is the sum of the areas of the rectangular region and the two semicircular regions, which is approximately (𝟗𝟗𝟔𝟔+ 𝟓𝟓𝟓𝟓.𝟏𝟏𝟏𝟏) 𝐢𝐢𝐢𝐢𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟔𝟔.𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏.

Common Mistake: Ask students to determine how to solve this problem and arrive at an incorrect solution of 196.48 in2. A student would arrive at this answer by including the area of the circle twice instead of once (50.24 in + 50.24 in + 96 in).

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If Marjorie wants to make six placemats, how many square inches of fabric will she need? Assume there is no waste.

There are 𝟔𝟔 placemats that are each 𝟏𝟏𝟏𝟏𝟔𝟔.𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏, so the fabric needed for all is 𝟔𝟔 ∙ 𝟏𝟏𝟏𝟏𝟔𝟔.𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏 = 𝟏𝟏𝟕𝟕𝟕𝟕.𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏.

Marjorie decides that she wants to sew on a contrasting band of material around the edge of the placemats. How much band material will Marjorie need?

The length of the band material needed will be the sum of the lengths of the two sides of the rectangular region and the circumference of the two semicircles (which is the same as the circumference of one circle with the same radius).

𝑷𝑷 = (𝒍𝒍 + 𝒍𝒍 + 𝟏𝟏𝝅𝝅𝒓𝒓)

𝑷𝑷 = (𝟏𝟏𝟏𝟏+ 𝟏𝟏𝟏𝟏 + 𝟏𝟏 ∙ 𝝅𝝅 ∙ 𝟏𝟏) = 𝟏𝟏𝟗𝟗.𝟏𝟏𝟏𝟏

The perimeter is 𝟏𝟏𝟗𝟗.𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏.


Example 3

The circumference of a circle is 𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐜𝐜. What is the exact area of the circle?

Draw a diagram to assist you in solving the problem.


The radius is needed to find the area of the circle. Let the radius be 𝒓𝒓 𝐜𝐜𝐜𝐜. Find the radius by using the circumference formula.

𝑪𝑪 = 𝟏𝟏𝝅𝝅𝒓𝒓 𝟏𝟏𝟏𝟏𝝅𝝅 = 𝟏𝟏𝝅𝝅𝒓𝒓

�𝟏𝟏𝟏𝟏𝝅𝝅

�𝟏𝟏𝟏𝟏𝝅𝝅 = �𝟏𝟏𝟏𝟏𝝅𝝅

�𝟏𝟏𝝅𝝅𝒓𝒓

𝟏𝟏𝟏𝟏 = 𝒓𝒓

The radius is 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜.

Next, find the area.

𝑨𝑨 = 𝝅𝝅 𝒓𝒓𝟏𝟏

𝑨𝑨 = 𝝅𝝅(𝟏𝟏𝟏𝟏)𝟏𝟏

𝑨𝑨 = 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅

The exact area of the circle is 𝟏𝟏𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐜𝐜𝟏𝟏.

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Exercises (10 minutes)

Students should solve these problems individually at first and then share with their cooperative groups after every other problem.

Exercises

1. Find the area of a circle with a diameter of 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜. Use 𝝅𝝅 ≈ 𝟏𝟏𝟏𝟏𝟕𝟕 .

If the diameter of the circle is 𝟏𝟏𝟏𝟏 𝒄𝒄𝒄𝒄, then the radius is 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 .

𝑨𝑨 = 𝝅𝝅𝒓𝒓𝟏𝟏

𝑨𝑨 ≈𝟏𝟏𝟏𝟏𝟕𝟕

(𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜)𝟏𝟏

𝑨𝑨 ≈ 𝟏𝟏𝟑𝟑𝟏𝟏𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏

2. The circumference of a circle is 𝟗𝟗𝝅𝝅 𝐜𝐜𝐜𝐜.

a. What is the diameter?

If 𝑪𝑪 = 𝝅𝝅𝝅𝝅, then 𝟗𝟗𝝅𝝅 𝐜𝐜𝐜𝐜 = 𝝅𝝅𝝅𝝅.

Solving the equation for the diameter, 𝝅𝝅, 𝟏𝟏𝝅𝝅∙ 𝟗𝟗𝝅𝝅 𝐜𝐜𝐜𝐜 =

𝟏𝟏𝝅𝝅𝝅𝝅 ∙ 𝝅𝝅.

So, 𝟗𝟗 𝐜𝐜𝐜𝐜 = 𝝅𝝅.

b. What is the radius?

If the diameter is 𝟗𝟗 𝐜𝐜𝐜𝐜, then the radius is half of that or 𝟗𝟗𝟏𝟏𝐜𝐜𝐜𝐜.

c. What is the area?

The area of the circle is 𝑨𝑨 = 𝝅𝝅 ∙ �𝟗𝟗𝟏𝟏 𝐜𝐜𝐜𝐜�𝟏𝟏, so 𝑨𝑨 = 𝟏𝟏𝟏𝟏

𝟏𝟏 𝝅𝝅 𝐜𝐜𝐜𝐜𝟏𝟏.

3. If students only know the radius of a circle, what other measures could they determine? Explain how students would use the radius to find the other parts.

If students know the radius, then they can find the diameter. The diameter is twice as long as the radius. The circumference can be found by doubling the radius and multiplying the result by 𝝅𝝅. The area can be found by multiplying the radius times itself and then multiplying that product by 𝝅𝝅.

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4. Find the area in the rectangle between the two quarter circles if 𝑨𝑨𝑨𝑨 = 𝟕𝟕 ft, 𝑨𝑨𝑭𝑭 = 𝟗𝟗 ft, and 𝑯𝑯𝑯𝑯 = 𝟕𝟕 ft. Use

𝝅𝝅 ≈ 𝟏𝟏𝟏𝟏𝟕𝟕 . Each quarter circle in the top-left and lower-right corners have the same radius.

The area between the quarter circles can be found by subtracting the area of the two quarter circles from the area of the rectangle. The area of the rectangle is the product of the length and the width. Side 𝑨𝑨𝑭𝑭 has a length of 𝟏𝟏𝟔𝟔 𝐟𝐟𝐟𝐟 and Side 𝑨𝑨𝑯𝑯 has a length of 𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟. The area of the rectangle is 𝑨𝑨 = 𝟏𝟏𝟔𝟔 𝐟𝐟𝐟𝐟 ∙ 𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟𝟏𝟏. The area of the two quarter circles is the same

as the area of a semicircle, which is half the area of a circle. 𝑨𝑨 = 𝟏𝟏𝟏𝟏 𝝅𝝅𝒓𝒓𝟏𝟏.

𝑨𝑨 ≈𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏𝟕𝟕∙ (𝟕𝟕 𝐟𝐟𝐟𝐟)𝟏𝟏

𝑨𝑨 ≈𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏𝟕𝟕∙ 𝟏𝟏𝟗𝟗 𝐟𝐟𝐟𝐟𝟏𝟏

𝑨𝑨 ≈ 𝟕𝟕𝟕𝟕 𝐟𝐟𝐟𝐟𝟏𝟏

The area between the two quarter circles is 𝟏𝟏𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟𝟏𝟏 − 𝟕𝟕𝟕𝟕 𝐟𝐟𝐟𝐟𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟕𝟕 𝐟𝐟𝐟𝐟𝟏𝟏.

Closing (5 minutes)

The area of a semicircular region is 12 of the area of a circle with the same radius.

The area of a quarter of a circular region is 14 of the area of a circle with the same radius.

If a problem asks you to use 227 for 𝜋𝜋, look for ways to use fraction arithmetic to simplify your computations in

the problem.

Problems that involve the composition of several shapes may be decomposed in more than one way.


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Name ___________________________________________________ Date____________________


Exit Ticket 1. Ken’s landscape gardening business creates odd-shaped lawns that include semicircles. Find the area of this

semicircular section of the lawn in this design. Use 227 for 𝜋𝜋.

2. In the figure below, Ken’s company has placed sprinkler heads at the center of the two small semicircles. The radius of the sprinklers is 5 ft. If the area in the larger semicircular area is the shape of the entire lawn, how much of the lawn will not be watered? Give your answer in terms of 𝜋𝜋 and to the nearest tenth. Explain your thinking.

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1. Ken’s landscape gardening business creates odd-shaped lawns that include semicircles. Find the area of this

semicircular section of the lawn in this design. Use 𝟏𝟏𝟏𝟏𝟕𝟕

for 𝝅𝝅.

If the diameter is 𝟓𝟓 𝐜𝐜, then the radius is 𝟓𝟓𝟏𝟏

𝐜𝐜. Using the formula for area of a semicircle,

𝑨𝑨 = 𝟏𝟏𝟏𝟏𝝅𝝅𝒓𝒓

𝟏𝟏, 𝑨𝑨 ≈ 𝟏𝟏𝟏𝟏 ∙

𝟏𝟏𝟏𝟏𝟕𝟕 ∙ �𝟓𝟓𝟏𝟏 𝐜𝐜�

𝟏𝟏. Using the order of operations,

𝑨𝑨 ≈ 𝟏𝟏𝟏𝟏 ∙

𝟏𝟏𝟏𝟏𝟕𝟕 ∙ 𝟏𝟏𝟓𝟓𝟏𝟏 𝐜𝐜𝟏𝟏 ≈ 𝟓𝟓𝟓𝟓𝟓𝟓

𝟓𝟓𝟔𝟔 𝐜𝐜𝟏𝟏 ≈ 𝟗𝟗.𝟏𝟏 𝐜𝐜𝟏𝟏.

2. In the figure below, Ken’s company has placed sprinkler heads at the center of the two small semicircles. The radius of the sprinklers is 𝟓𝟓 𝐟𝐟𝐟𝐟. If the area in the larger semicircular area is the shape of the entire lawn, how much of the lawn will not be watered? Give your answer in terms of 𝝅𝝅 and to the nearest tenth. Explain your thinking.

The area not covered by the sprinklers would be the area between the larger semicircle and the two smaller ones. The area for the two semicircles is the same as the area of one circle with the same radius of 𝟓𝟓 ft. The area not covered by the sprinklers can be found by subtracting the area of the two smaller semicircles from the area of the large semicircle.

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 𝐍𝐍𝐍𝐍𝐟𝐟 𝐂𝐂𝐍𝐍𝐂𝐂𝐀𝐀𝐀𝐀𝐀𝐀𝐂𝐂 = 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 𝐍𝐍𝐟𝐟 𝐥𝐥𝐀𝐀𝐀𝐀𝐥𝐥𝐀𝐀 𝐬𝐬𝐀𝐀𝐜𝐜𝐢𝐢𝐜𝐜𝐢𝐢𝐀𝐀𝐜𝐜𝐥𝐥𝐀𝐀 − 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 𝐍𝐍𝐟𝐟 𝐟𝐟𝐭𝐭𝐍𝐍 𝐬𝐬𝐜𝐜𝐀𝐀𝐥𝐥𝐥𝐥𝐀𝐀𝐀𝐀 𝐬𝐬𝐀𝐀𝐜𝐜𝐢𝐢𝐜𝐜𝐢𝐢𝐀𝐀𝐜𝐜𝐥𝐥𝐀𝐀𝐬𝐬

𝑨𝑨 = 𝟏𝟏𝟏𝟏𝝅𝝅 ∙ (𝟏𝟏𝟓𝟓 𝐟𝐟𝐟𝐟)𝟏𝟏 − �𝟏𝟏 ∙ �

𝟏𝟏𝟏𝟏

(𝝅𝝅 ∙ (𝟓𝟓 𝐟𝐟𝐟𝐟)𝟏𝟏)��

𝑨𝑨 = 𝟏𝟏𝟏𝟏𝝅𝝅 ∙ 𝟏𝟏𝟓𝟓𝟓𝟓 𝐟𝐟𝐟𝐟𝟏𝟏 − 𝝅𝝅 ∙ 𝟏𝟏𝟓𝟓 𝐟𝐟𝐟𝐟𝟏𝟏

𝑨𝑨 = 𝟓𝟓𝟓𝟓𝝅𝝅 𝐟𝐟𝐟𝐟𝟏𝟏 − 𝟏𝟏𝟓𝟓𝝅𝝅 𝐟𝐟𝐟𝐟𝟏𝟏 = 𝟏𝟏𝟓𝟓𝝅𝝅 𝐟𝐟𝐟𝐟𝟏𝟏

Let 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏

𝑨𝑨 ≈ 𝟕𝟕𝟏𝟏.𝟓𝟓 𝐟𝐟𝐟𝐟𝟏𝟏

The sprinklers will not cover 𝟏𝟏𝟓𝟓𝝅𝝅 𝐟𝐟𝐟𝐟𝟏𝟏 or 𝟕𝟕𝟏𝟏.𝟓𝟓 𝐟𝐟𝐟𝐟𝟏𝟏 of the lawn.


1. Mark created a flower bed that is semicircular in shape. The diameter of the flower bed is 𝟓𝟓 𝐜𝐜 .

a. What is the perimeter of the flower bed? (Approximate 𝝅𝝅 to be 𝟑𝟑.𝟏𝟏𝟏𝟏.)

The perimeter of this flower bed is the sum of the diameter and one-half the circumference of a circle with the same diameter.

𝑷𝑷 = 𝐂𝐂𝐢𝐢𝐀𝐀𝐜𝐜𝐀𝐀𝐟𝐟𝐀𝐀𝐀𝐀+𝟏𝟏𝟏𝟏𝝅𝝅 ∙ 𝐂𝐂𝐢𝐢𝐀𝐀𝐜𝐜𝐀𝐀𝐟𝐟𝐀𝐀𝐀𝐀

𝑷𝑷 ≈ 𝟓𝟓 𝐜𝐜+ 𝟏𝟏𝟏𝟏∙ 𝟑𝟑.𝟏𝟏𝟏𝟏 ∙ 𝟓𝟓 𝐜𝐜

𝑷𝑷 ≈ 𝟏𝟏𝟏𝟏.𝟏𝟏𝟓𝟓 𝐜𝐜

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b. What is the area of the flower bed? (Approximate 𝝅𝝅 to be 𝟑𝟑.𝟏𝟏𝟏𝟏.)

𝑨𝑨 =𝟏𝟏𝟏𝟏𝝅𝝅 (𝟏𝟏.𝟓𝟓 𝐜𝐜)𝟏𝟏

𝑨𝑨 =𝟏𝟏𝟏𝟏𝝅𝝅 (𝟔𝟔.𝟏𝟏𝟓𝟓 𝐜𝐜𝟏𝟏)

𝑨𝑨 ≈ 𝟓𝟓.𝟓𝟓 ∙ 𝟑𝟑.𝟏𝟏𝟏𝟏 ∙ 𝟔𝟔.𝟏𝟏𝟓𝟓 𝐜𝐜𝟏𝟏 𝑨𝑨 ≈ 𝟗𝟗.𝟏𝟏 𝐜𝐜𝟏𝟏

2. A landscape designer wants to include a semicircular patio at the end of a square sandbox. She knows that the area of the semicircular patio is 𝟏𝟏𝟓𝟓.𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏.

a. Draw a picture to represent this situation.

b. What is the length of the side of the square?

If the area of the patio is 𝟏𝟏𝟓𝟓.𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏, then we can find the radius by solving the equation 𝑨𝑨 = 𝟏𝟏𝟏𝟏𝝅𝝅𝒓𝒓

𝟏𝟏 and substituting the information that we know. If we approximate 𝝅𝝅 to be 𝟑𝟑.𝟏𝟏𝟏𝟏 and solve for the radius, 𝒓𝒓, then

𝟏𝟏𝟓𝟓.𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏 ≈𝟏𝟏𝟏𝟏𝝅𝝅𝒓𝒓𝟏𝟏

𝟏𝟏𝟏𝟏∙ 𝟏𝟏𝟓𝟓.𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏 ≈

𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏𝝅𝝅𝒓𝒓𝟏𝟏

𝟓𝟓𝟓𝟓.𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏𝒓𝒓𝟏𝟏 𝟏𝟏

𝟑𝟑.𝟏𝟏𝟏𝟏∙ 𝟓𝟓𝟓𝟓.𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏 ≈

𝟏𝟏𝟑𝟑.𝟏𝟏𝟏𝟏

∙ 𝟑𝟑.𝟏𝟏𝟏𝟏𝒓𝒓𝟏𝟏

𝟏𝟏𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏 ≈ 𝒓𝒓𝟏𝟏 𝟏𝟏 𝐜𝐜𝐜𝐜 ≈ 𝒓𝒓

The length of the diameter is 𝟏𝟏 𝐜𝐜𝐜𝐜; therefore, the length of the side of the square is 𝟏𝟏 𝐜𝐜𝐜𝐜.

3. A window manufacturer designed a set of windows for the top of a two-story wall. If the window is comprised of 𝟏𝟏 squares and 𝟏𝟏 quarter circles on each end, and if the length of the span of windows across the bottom is 𝟏𝟏𝟏𝟏 feet, approximately how much glass will be needed to complete the set of windows?

The area of the windows is the sum of the areas of the two quarter circles and the two squares that make up the

bank of windows. If the span of windows is 𝟏𝟏𝟏𝟏 feet across the bottom, then each window is 𝟑𝟑 feet wide on the

bottom. The radius of the quarter circles is 𝟑𝟑 feet, so the area for one quarter circle window is 𝑨𝑨 = 𝟏𝟏𝟏𝟏𝝅𝝅 ∙ (𝟑𝟑 𝐟𝐟𝐟𝐟)𝟏𝟏, or

𝑨𝑨 ≈ 𝟕𝟕.𝟓𝟓𝟔𝟔𝟓𝟓 𝐟𝐟𝐟𝐟𝟏𝟏. The area of one square window is 𝑨𝑨 = (𝟑𝟑 𝐟𝐟𝐟𝐟)𝟏𝟏, or 𝟗𝟗 𝐟𝐟𝐟𝐟𝟏𝟏. The total area is 𝑨𝑨 =𝟏𝟏(𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 𝐍𝐍𝐟𝐟 𝐪𝐪𝐪𝐪𝐀𝐀𝐀𝐀𝐟𝐟𝐀𝐀𝐀𝐀 𝐜𝐜𝐢𝐢𝐀𝐀𝐜𝐜𝐥𝐥𝐀𝐀) + 𝟏𝟏(𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 𝐍𝐍𝐟𝐟 𝐬𝐬𝐪𝐪𝐪𝐪𝐀𝐀𝐀𝐀𝐀𝐀), or 𝑨𝑨 ≈ (𝟏𝟏 ∙ 𝟕𝟕.𝟓𝟓𝟔𝟔𝟓𝟓 𝐟𝐟𝐟𝐟𝟏𝟏) + (𝟏𝟏 ∙ 𝟗𝟗 𝐟𝐟𝐟𝐟𝟏𝟏) ≈ 𝟑𝟑𝟏𝟏.𝟏𝟏𝟑𝟑 𝐟𝐟𝐟𝐟𝟏𝟏.

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4. Find the area of the shaded region. (Approximate 𝝅𝝅 to be 𝟏𝟏𝟏𝟏𝟕𝟕

.)

5. The figure below shows a circle inside of a square. If the radius of the circle is 𝟏𝟏 𝐜𝐜𝐜𝐜, find the following and explain your solution.

a. The circumference of the circle

𝑪𝑪 = 𝟏𝟏𝝅𝝅 ∙ 𝟏𝟏 𝐜𝐜𝐜𝐜

𝑪𝑪 = 𝟏𝟏𝟔𝟔𝝅𝝅 𝐜𝐜𝐜𝐜

b. The area of the circle

𝑨𝑨 = 𝝅𝝅 ∙ (𝟏𝟏 𝐜𝐜𝐜𝐜)𝟏𝟏

𝑨𝑨 = 𝟔𝟔𝟏𝟏 𝝅𝝅 𝐜𝐜𝐜𝐜𝟏𝟏

c. The area of the square

𝑨𝑨 = 𝟏𝟏𝟔𝟔 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏𝟔𝟔 𝐜𝐜𝐜𝐜

𝑨𝑨 = 𝟏𝟏𝟓𝟓𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏

6. Michael wants to create a tile pattern out of three quarter circles for his kitchen backsplash. He will repeat the three quarter circles throughout the pattern. Find the area of the tile pattern that Michael will use. Approximate 𝝅𝝅 as 𝟑𝟑.𝟏𝟏𝟏𝟏.

There are three quarter circles in the tile design. The area of one quarter circle multiplied by 𝟑𝟑 will result in the total area.

𝑨𝑨 =𝟏𝟏𝟏𝟏𝝅𝝅 ∙ (𝟏𝟏𝟔𝟔 𝐜𝐜𝐜𝐜)𝟏𝟏

𝑨𝑨 ≈𝟏𝟏𝟏𝟏∙ 𝟑𝟑.𝟏𝟏𝟏𝟏 ∙ 𝟏𝟏𝟓𝟓𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏

𝑨𝑨 ≈ 𝟏𝟏𝟓𝟓𝟓𝟓.𝟗𝟗𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏

𝑨𝑨 ≈ 𝟑𝟑 ∙ 𝟏𝟏𝟓𝟓𝟓𝟓.𝟗𝟗𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏

𝑨𝑨 ≈ 𝟔𝟔𝟓𝟓𝟏𝟏.𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏

The area of the tile pattern is approximately 𝟔𝟔𝟓𝟓𝟏𝟏.𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏.

𝑨𝑨 =𝟏𝟏𝟏𝟏𝝅𝝅(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢)𝟏𝟏

𝑨𝑨 =𝟏𝟏𝟏𝟏𝝅𝝅 ∙ 𝟏𝟏𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏

𝑨𝑨 ≈𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏𝟕𝟕∙ 𝟏𝟏𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏

𝑨𝑨 ≈𝟕𝟕𝟗𝟗𝟏𝟏𝟕𝟕

𝐢𝐢𝐢𝐢𝟏𝟏 or 𝟏𝟏𝟏𝟏𝟑𝟑.𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏

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7. A machine shop has a square metal plate with sides that measure 𝟏𝟏 𝐜𝐜𝐜𝐜 each. A machinist must cut four semicircles

with a radius of 𝟏𝟏𝟏𝟏𝐜𝐜𝐜𝐜 and four quarter circles with a radius of 𝟏𝟏 cm from its sides and corners. What is the area of

the plate formed? Use 𝟏𝟏𝟏𝟏𝟕𝟕

to approximate 𝝅𝝅.

8. A graphic artist is designing a company logo with two concentric circles (two circles that share the same center but have different radii). The artist needs to know the area of the shaded band between the two concentric circles. Explain to the artist how he would go about finding the area of the shaded region.

9. Create your own shape made up of rectangles, squares, circles, or semicircles, and determine the area and perimeter.

Student answers may vary.

The area of the metal plate is determined by subtracting the four quarter circles (corners) and the four half-circles (on each side) from the area of the square. Area of the square: 𝑨𝑨 = (𝟏𝟏 𝐜𝐜𝐜𝐜)𝟏𝟏 = 𝟏𝟏𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏.

The area of four quarter circles is the same as the area of a circle with a radius of

𝟏𝟏 𝐜𝐜𝐜𝐜: 𝑨𝑨 ≈ 𝟏𝟏𝟏𝟏𝟕𝟕 (𝟏𝟏 𝐜𝐜𝐜𝐜)𝟏𝟏 ≈ 𝟏𝟏𝟏𝟏

𝟕𝟕 𝐜𝐜𝐜𝐜𝟏𝟏.

The area of the four semicircles with radius 𝟏𝟏𝟏𝟏𝐜𝐜𝐜𝐜 is

𝑨𝑨 ≈ 𝟏𝟏 ∙𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏𝟕𝟕∙ �𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜�𝟏𝟏

𝑨𝑨 ≈ 𝟏𝟏 ∙𝟏𝟏𝟏𝟏∙𝟏𝟏𝟏𝟏𝟕𝟕∙𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜𝟏𝟏 ≈𝟏𝟏𝟏𝟏𝟕𝟕𝐜𝐜𝐜𝐜𝟏𝟏.

The area of the metal plate is

𝑨𝑨 ≈ 𝟏𝟏𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏 −𝟏𝟏𝟏𝟏𝟕𝟕

𝐜𝐜𝐜𝐜𝟏𝟏 −𝟏𝟏𝟏𝟏𝟕𝟕

𝐜𝐜𝐜𝐜𝟏𝟏 ≈ 𝟕𝟕𝟗𝟗𝟕𝟕

𝐜𝐜𝐜𝐜𝟏𝟏

The artist should find the areas of both the larger and smaller circles. Then, the artist should subtract the area of the smaller circle from the area of the larger circle to find the area between the two circles. The area of the larger circle is

𝑨𝑨 = 𝝅𝝅 ∙ (𝟗𝟗 𝐜𝐜𝐜𝐜)𝟏𝟏 or 𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐜𝐜𝟏𝟏.

The area of the smaller circle is

𝑨𝑨 = 𝝅𝝅(𝟓𝟓 𝐜𝐜𝐜𝐜)𝟏𝟏 or 𝟏𝟏𝟓𝟓𝝅𝝅 𝐜𝐜𝐜𝐜𝟏𝟏.

The area of the region between the circles is 𝟏𝟏𝟏𝟏𝝅𝝅 𝐜𝐜𝐜𝐜𝟏𝟏 − 𝟏𝟏𝟓𝟓𝟐𝟐 𝐜𝐜𝐜𝐜𝟏𝟏 = 𝟓𝟓𝟔𝟔𝝅𝝅 𝐜𝐜𝐜𝐜𝟏𝟏. If we approximate 𝝅𝝅 to be 𝟑𝟑.𝟏𝟏𝟏𝟏, then 𝑨𝑨 ≈ 𝟏𝟏𝟕𝟕𝟓𝟓.𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟏𝟏.

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7•3 Lesson 19

Lesson 19: Unknown Area Problems on the Coordinate Plane

Lesson 19: Unknown Area Problems on the Coordinate

Plane

Student Outcomes

Students find the areas of triangles and simple polygonal regions in the coordinate plane with vertices at grid points by composing into rectangles and decomposing into triangles and quadrilaterals.

Lesson Notes Students extend their knowledge of finding area to figures on a coordinate plane. The lesson begins with a proof of the area of a parallelogram. In Grade 6, students proved the area of a parallelogram through a different approach. This lesson draws heavily on MP.7 (look for and make use of structure). Students notice and take advantage of figures composed of simpler ones to determine area.

Classwork

Example (20 minutes): Area of a Parallelogram

Allow students to work through parts (a)–(e) of the example either independently or in groups. Circulate around the room to check student progress and to ensure that students are drawing the figures correctly. Debrief before having them move on to part (f).

Example: Area of a Parallelogram

The coordinate plane below contains figure 𝑷𝑷, parallelogram 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨.

a. Write the ordered pairs of each of the vertices next to the vertex points.

See figure.

b. Draw a rectangle surrounding figure 𝑷𝑷 that has vertex points of 𝑨𝑨 and 𝑨𝑨. Label the two triangles in the figure as 𝑺𝑺 and 𝑻𝑻.

See figure.

c. Find the area of the rectangle.

Base = 𝟖𝟖 units

Height = 𝟔𝟔 units

Area = 𝟖𝟖 units × 𝟔𝟔 units = 𝟒𝟒𝟖𝟖 sq. units

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d. Find the area of each triangle.

Figure 𝑺𝑺 Figure 𝑻𝑻

Base = 𝟑𝟑 units Base = 𝟑𝟑 units

Height = 𝟔𝟔 units Height = 𝟔𝟔 units

Area= 𝟏𝟏𝟐𝟐

× 𝟑𝟑 units × 𝟔𝟔 units Area = 𝟏𝟏𝟐𝟐

× 𝟑𝟑 units × 𝟔𝟔 units

= 𝟗𝟗 sq. units = 𝟗𝟗 sq. units

e. Use these areas to find the area of parallelogram 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨.

Area 𝑷𝑷 = Area of rectangle − Area 𝑺𝑺 − Area 𝑻𝑻

= 𝟒𝟒𝟖𝟖 sq. units − 𝟗𝟗 sq. units − 𝟗𝟗 sq. units = 𝟑𝟑𝟑𝟑 sq. units

Stop students here and discuss responses.

How did you find the base and height of each figure?

By using the scale on the coordinate plane

How did you find the area of the parallelogram?

By subtracting the areas of the triangles from the area of the rectangle

Assist students with part (f) if necessary and then give them time to finish the exploration.

The coordinate plane below contains figure 𝑹𝑹, a rectangle with the same base as the parallelogram above.

f. Draw triangles 𝑺𝑺 and 𝑻𝑻 and connect to figure 𝑹𝑹 so that you create a rectangle that is the same size as the rectangle you created on the first coordinate plane.

See figure.

g. Find the area of rectangle 𝑹𝑹.

Base = 𝟓𝟓 units

Height = 𝟔𝟔 units

Area = 𝟑𝟑𝟑𝟑 sq. units

h. What do figures 𝑹𝑹 and 𝑷𝑷 have in common?

They have the same area. They share the same base and have the same height.

Debrief and allow students to share responses. Draw the height of the parallelogram to illustrate that it has the same height as rectangle 𝑅𝑅.

Since the larger rectangles are the same size, their areas must be equal. Write this on the board:

Area of 𝑃𝑃 + Area of 𝑆𝑆 + Area of 𝑇𝑇 = Area of 𝑅𝑅 + Area of 𝑆𝑆 + Area of 𝑇𝑇

Based on the equation, what must be true about the area of 𝑃𝑃?

Area of 𝑃𝑃 = Area of 𝑅𝑅 How can we find the area of a parallelogram?

Area of Parallelogram = base × height

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Exercises (17 minutes)

Have students work on the exercises independently and then check answers with a partner. Then, discuss results as a class.

Exercises

1. Find the area of triangle 𝑨𝑨𝑨𝑨𝑨𝑨.

𝑨𝑨 =𝟏𝟏𝟐𝟐

× 𝟕𝟕 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 × 𝟒𝟒 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 = 𝟏𝟏𝟒𝟒 𝐮𝐮𝐬𝐬.𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮

2. Find the area of quadrilateral 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 two different ways.

𝟏𝟏𝟐𝟐

× 𝟐𝟐 × 𝟓𝟓+ 𝟐𝟐 × 𝟓𝟓 +𝟏𝟏𝟐𝟐

× 𝟏𝟏× 𝟓𝟓 = 𝟓𝟓 + 𝟏𝟏𝟑𝟑+ 𝟐𝟐.𝟓𝟓 = 𝟏𝟏𝟕𝟕.𝟓𝟓

The area is 𝟏𝟏𝟕𝟕.𝟓𝟓 𝐮𝐮𝐬𝐬.𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮.

𝟏𝟏𝟐𝟐

× (𝟓𝟓+ 𝟐𝟐) × 𝟓𝟓 = 𝟏𝟏𝟕𝟕.𝟓𝟓

The area is 𝟏𝟏𝟕𝟕.𝟓𝟓 𝐮𝐮𝐬𝐬.𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮.

3. The area of quadrilateral 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 is 𝟏𝟏𝟐𝟐 sq. units. Find 𝒙𝒙.

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝐛𝐛𝐀𝐀𝐮𝐮𝐀𝐀× 𝐡𝐡𝐀𝐀𝐮𝐮𝐡𝐡𝐡𝐡𝐮𝐮 𝟏𝟏𝟐𝟐 𝐮𝐮𝐬𝐬.𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 = 𝟐𝟐𝒙𝒙

𝟔𝟔 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 = 𝒙𝒙

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4. The area of triangle 𝑨𝑨𝑨𝑨𝑨𝑨 is 𝟏𝟏𝟒𝟒 sq. units. Find the length of side 𝑨𝑨𝑨𝑨��.

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 =𝟏𝟏𝟐𝟐

× 𝐛𝐛𝐀𝐀𝐮𝐮𝐀𝐀× 𝐡𝐡𝐀𝐀𝐮𝐮𝐡𝐡𝐡𝐡𝐮𝐮

𝟏𝟏𝟒𝟒 𝐮𝐮𝐬𝐬.𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 =𝟏𝟏𝟐𝟐

× 𝑨𝑨𝑨𝑨 × (𝟕𝟕 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮)

𝑨𝑨𝑨𝑨 = 𝟒𝟒 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮


Area of rectangle 𝑨𝑨𝑹𝑹𝑺𝑺𝑻𝑻 = 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 × 𝟏𝟏𝟑𝟑 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 = 𝟏𝟏𝟏𝟏𝟑𝟑 𝐮𝐮𝐬𝐬.𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮

Area of triangle 𝑨𝑨𝑹𝑹𝑨𝑨 = 𝟏𝟏𝟐𝟐

× 𝟕𝟕 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 × 𝟏𝟏𝟑𝟑 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 = 𝟑𝟑𝟓𝟓 𝐮𝐮𝐬𝐬. 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮

Area of triangle 𝑨𝑨𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟐𝟐

× 𝟒𝟒 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 × 𝟓𝟓 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 = 𝟏𝟏𝟑𝟑 𝐮𝐮𝐬𝐬. 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮

Area of triangle 𝑨𝑨𝑻𝑻𝑨𝑨 = 𝟏𝟏𝟐𝟐

× 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 × 𝟓𝟓 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 = 𝟐𝟐𝟕𝟕. 𝟓𝟓 𝐮𝐮𝐬𝐬. 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮

Area of triangle 𝑨𝑨𝑨𝑨𝑨𝑨 = Area of 𝑨𝑨𝑹𝑹𝑺𝑺𝑻𝑻− Area of 𝑨𝑨𝑹𝑹𝑨𝑨− Area of 𝑨𝑨𝑺𝑺𝑨𝑨 − Area of 𝑨𝑨𝑻𝑻𝑨𝑨 = 𝟑𝟑𝟕𝟕.𝟓𝟓 𝐮𝐮𝐬𝐬.𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮

What shape is the quadrilateral in Exercise 2?

Trapezoid

What methods did you use to find the area?

Decomposing the figure into two right triangles and a rectangle or using the area formula for a trapezoid

Which method was easier for finding the area?

Answers will vary.

For Exercise 4, what piece of information was missing? Why couldn’t we find it using the coordinate plane?

The base was missing. We could measure the height but not the base because no scale was given on the 𝑥𝑥-axis.

For Exercise 5, why couldn’t we find the area of triangle 𝐴𝐴𝐴𝐴𝐴𝐴 by simply using its base and height?

Because of the way the triangle was oriented, we could not measure the exact length of the base or the height using the coordinate plane.

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7•3 Lesson 19


Closing (3 minutes)

Review relevant vocabulary and formulas from this lesson. These terms and formulas should be a review from earlier grades and previous lessons in this module.

Relevant Vocabulary:

Quadrilateral Parallelogram Trapezoid

Rectangle Square Altitude and base of a triangle

Semicircle Diameter of a circle

Area formulas:

Area of parallelogram = base × height Area of rectangle = base × height

Area of a triangle = 12

× base × height Area of a trapezoid = 12

× (base 1 + base 2) × height

Area of a circle = 𝜋𝜋 × 𝑟𝑟2

Why is it useful to have a figure on a coordinate plane?

The scale can be used to measure the base and height.

What are some methods for finding the area of a quadrilateral?

Use a known area formula, deconstruct the figure into shapes with known area formulas, make the figure a part of a larger shape and then subtract areas.


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Name Date


Exit Ticket The figure 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 is a rectangle. 𝐴𝐴𝐴𝐴 = 2 units, 𝐴𝐴𝐴𝐴 = 4 units, and 𝐴𝐴𝐴𝐴 = 𝐹𝐹𝐴𝐴 = 1 unit.

1. Find the area of rectangle 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴.

2. Find the area of triangle 𝐴𝐴𝐴𝐴𝐴𝐴.

3. Find the area of triangle 𝐴𝐴𝐴𝐴𝐹𝐹.

4. Find the area of the parallelogram 𝐴𝐴𝐴𝐴𝐴𝐴𝐹𝐹 two different ways.

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The figure 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 is a rectangle. 𝑨𝑨𝑨𝑨 = 𝟐𝟐 units, 𝑨𝑨𝑨𝑨 = 𝟒𝟒 units, and 𝑨𝑨𝑨𝑨 = 𝑭𝑭𝑨𝑨 = 𝟏𝟏 unit.

1. Find the area of rectangle 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨.

Area = 𝟒𝟒 units × 𝟐𝟐 units = 𝟖𝟖 sq. units


Area = 𝟏𝟏𝟐𝟐 × 𝟏𝟏 unit × 𝟐𝟐 units = 𝟏𝟏 sq. unit

3. Find the area of triangle 𝑨𝑨𝑨𝑨𝑭𝑭.

Area = 𝟏𝟏𝟐𝟐 × 𝟏𝟏 unit × 𝟐𝟐 units = 𝟏𝟏 sq. unit

4. Find the area of the parallelogram 𝑨𝑨𝑨𝑨𝑨𝑨𝑭𝑭 two different ways.

Area = Area of 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 – Area of 𝑨𝑨𝑨𝑨𝑨𝑨 – Area of 𝑨𝑨𝑨𝑨𝑭𝑭

= (𝟖𝟖 − 𝟏𝟏 − 𝟏𝟏) sq. units = 𝟔𝟔 sq. units

Area = base × height

= 𝟑𝟑 units × 𝟐𝟐 units = 𝟔𝟔 sq. units


Find the area of each figure.

1.

Area = 𝟏𝟏𝟑𝟑.𝟓𝟓 sq. units

2.

Area = 𝟒𝟒.𝟓𝟓𝝅𝝅 sq. units ≈ 𝟏𝟏𝟒𝟒.𝟏𝟏𝟑𝟑 sq. units

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3.

Area = 𝟒𝟒𝟖𝟖 sq. units

4.

Area = (𝟐𝟐𝝅𝝅+ 𝟏𝟏𝟔𝟔) sq. units ≈ 𝟐𝟐𝟐𝟐.𝟐𝟐𝟖𝟖 sq. units

5.

Area = 𝟔𝟔𝟖𝟖 sq. units

6.

Area = 𝟒𝟒𝟔𝟔 sq. units

For Problems 7–9, draw a figure in the coordinate plane that matches each description.

7. A rectangle with an area of 𝟏𝟏𝟖𝟖 sq. units

8. A parallelogram with an area of 𝟓𝟓𝟑𝟑 sq. units

9. A triangle with an area of 𝟐𝟐𝟓𝟓 sq. units

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Find the unknown value labled as 𝒙𝒙 on each figure.

10. The rectangle has an area of 𝟖𝟖𝟑𝟑 sq. units.

𝒙𝒙 = 𝟖𝟖

11. The trapezoid has an area of 𝟏𝟏𝟏𝟏𝟓𝟓 sq. units.

𝒙𝒙 = 𝟏𝟏𝟑𝟑


Area = 𝟔𝟔.𝟓𝟓 sq. units

13. Find the area of the quadrilateral using two different methods. Describe the methods used, and explain why they result in the same area.

Area = 𝟏𝟏𝟓𝟓 sq. units

One method is by drawing a rectangle around the figure. The area of the parallelogram is equal to the area of the rectangle minus the area of the two triangles. A second method is to use the area formula for a parallelogram (Area = base × height).

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14. Find the area of the quadrilateral using two different methods. What are the advantages or disadvantages of each method?

Area = 𝟔𝟔𝟑𝟑 sq. units

One method is to use the area formula for a trapezoid, 𝑨𝑨 = 𝟏𝟏𝟐𝟐 (𝐛𝐛𝐀𝐀𝐮𝐮𝐀𝐀 𝟏𝟏+ 𝐛𝐛𝐀𝐀𝐮𝐮𝐀𝐀 𝟐𝟐) × 𝐡𝐡𝐀𝐀𝐮𝐮𝐡𝐡𝐡𝐡𝐮𝐮. The second method is

to split the figure into a rectangle and a triangle. The second method requires more calculations. The first method requires first recognizing the figure as a trapezoid and recalling the formula for the area of a trapezoid.

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7•3 Lesson 20

Lesson 20: Composite Area Problems


Student Outcomes

Students find the area of regions in the coordinate plane with polygonal boundaries by decomposing the plane into triangles and quadrilaterals, including regions with polygonal holes.

Students find composite areas of regions in the coordinate plane by decomposing the plane into familiar figures (triangles, quadrilaterals, circles, semicircles, and quarter circles).

Lesson Notes In Lessons 17 through 20, students learned to find the areas of various regions, including quadrilaterals, triangles, circles, semicircles, and those plotted on coordinate planes. In this lesson, students use prior knowledge to use the sum and/or difference of the areas to find unknown composite areas.

Classwork


Example 1

Find the composite area of the shaded region. Use 𝟑𝟑.𝟏𝟏𝟏𝟏 for 𝝅𝝅.

Allow students to look at the problem and find the area independently before solving as a class.

What information can we take from the image? Two circles are on the coordinate plane. The diameter of the larger circle is 6 units, and the diameter of

the smaller circle is 4 units.

How do we know what the diameters of the circles are?

We can count the units along the diameter of the circles, or we can subtract the coordinate points to find the length of the diameter.

Scaffolding: For struggling students, display posters around the room displaying the visuals and the formulas of the area of a circle, a triangle, and a quadrilateral for reference.

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7•3 Lesson 20


What information do we know about circles?

The area of a circle is equal to the radius squared times 𝜋𝜋. We can approximate 𝜋𝜋 as 3.14 or 227

.

After calculating the two areas, what is the next step, and how do you know?

The non-overlapping regions add, meaning that the Area(small disk) + Area(ring) = Area(big disk). Rearranging this results in this: Area(ring) = Area(big disk) − Area(small disk). So, the next step is to take the difference of the disks.

What is the area of the figure?

9𝜋𝜋 − 4𝜋𝜋 = 5𝜋𝜋; the area of the figure is approximately 15.7 square units.


A yard is shown with the shaded section indicating grassy areas and the unshaded sections indicating paved areas. Find the area of the space covered with grass in units2.

Area of rectangle 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 − area of rectangle 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 = area of shaded region

(𝟑𝟑 ∙ 𝟐𝟐) − �𝟏𝟏𝟐𝟐∙ 𝟏𝟏�

𝟔𝟔 −𝟏𝟏𝟐𝟐

𝟓𝟓𝟏𝟏𝟐𝟐

The area of the space covered with grass is 𝟓𝟓𝟏𝟏𝟐𝟐 units².

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Example 2

Find the area of the figure that consists of a rectangle with a semicircle on top. Use 𝟑𝟑.𝟏𝟏𝟏𝟏 for 𝝅𝝅.

𝑨𝑨 = 𝟐𝟐𝟐𝟐.𝟐𝟐𝟐𝟐 𝒎𝒎𝟐𝟐

What do we know from reading the problem and looking at the picture?

There is a semicircle and a rectangle.

What information do we need to find the areas of the circle and the rectangle? We need to know the base and height of the rectangle and the radius

of the semicircle. For this problem, let the radius for the semicircle be 𝑟𝑟 meters.

How do we know where to draw the diameter of the circle?

The diameter is parallel to the bottom base of the rectangle because we know that the figure includes a semicircle.

What is the diameter and radius of the circle?

The diameter of the circle is equal to the base of the rectangle, 4 m . The radius is half of 4 m, which is 2 m.

What would a circle with a diameter of 4 m look like relative to the figure?

What is the importance of labeling the known lengths of the figure?

This helps us keep track of the lengths when we need to use them to calculate different parts of the composite figure. It also helps us find unknown lengths because they may be the sum or the difference of known lengths.

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How do we find the base and height of the rectangle?

The base is labeled 4 m, but the height of the rectangle is combined with the radius of the semicircle. The difference of the height of the figure, 7.5 m, and the radius of the semicircle equals the height of the rectangle. Thus, the height of the rectangle is (7.5 − 2) m, which equals 5.5 m.

What is the area of the rectangle?

The area of the rectangle is 5.5 m times 4 m. The area is 22.0 m2.

What is the area of the semicircle?

The area of the semicircle is half the area of a circle with a radius of 2 m. The area is 4(3.14) m2 divided by 2, which equals 6.28 m2.

Do we subtract these areas as we did in Example 1?

No, we combine the two. The figure is the sum of the rectangle and the semicircle.

What is the area of the figure?

28.28 m2


Students work in pairs to decompose the figure into familiar shapes and find the area.

Exercise 2

Find the area of the shaded region. Use 𝟑𝟑.𝟏𝟏𝟏𝟏 for 𝝅𝝅.

Area of the triangle + area of the semicircle = area of the shaded region

�𝟏𝟏𝟐𝟐

𝒃𝒃 × 𝒉𝒉� + �𝟏𝟏𝟐𝟐� (𝝅𝝅𝒓𝒓𝟐𝟐)

�𝟏𝟏𝟐𝟐∙ 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 ∙ 𝟐𝟐 𝐜𝐜𝐜𝐜�+ �

𝟏𝟏𝟐𝟐� (𝟑𝟑.𝟏𝟏𝟏𝟏 ∙ (𝟏𝟏 𝐜𝐜𝐜𝐜)𝟐𝟐)

𝟓𝟓𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟐𝟐𝟓𝟓.𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐

𝟐𝟐𝟏𝟏.𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐

The area is approximately 𝟐𝟐𝟏𝟏.𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐.

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Using the figure below, have students work in pairs to create a plan to find the area of the shaded region and to label known values. Emphasize to students that they should label known lengths to assist in finding the areas. Reconvene as a class to discuss the possible ways of finding the area of the shaded region. Discern which discussion questions to address depending on the level of students.

Example 3

Find the area of the shaded region.

What recognizable shapes are in the figure?

A square and a triangle

What else is created by these two shapes?

There are three right triangles. What specific shapes comprise the square?

Three right triangles and one non-right triangle

Redraw the figure separating the triangles; then, label the lengths discussing the calculations.

Do we know any of the lengths of the acute triangle?

No

Do we have information about the right triangles? Yes, because of the given lengths, we can calculate unknown sides.

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Is the sum or difference of these parts needed to find the area of the shaded region?

Both are needed. The difference of the square and the sum of the three right triangles is the area of the shaded triangle.

What is the area of the shaded region?

400 cm2 − ��12

× 20 cm × 12 cm� + �12

× 20 cm × 14 cm� + �12

× 8 cm × 6 cm�� = 116 cm2

The area is 116 cm².


Find the area of the shaded region. The figure is not drawn to scale.

Area of squares – (area of the bottom right triangle + area of the top right triangle)

�(𝟐𝟐 𝐜𝐜𝐜𝐜 × 𝟐𝟐 𝐜𝐜𝐜𝐜) + (𝟑𝟑 𝐜𝐜𝐜𝐜× 𝟑𝟑 𝐜𝐜𝐜𝐜)� − ��𝟏𝟏𝟐𝟐

× 𝟓𝟓 𝐜𝐜𝐜𝐜 × 𝟐𝟐 𝐜𝐜𝐜𝐜�+ �𝟏𝟏𝟐𝟐

× 𝟑𝟑 𝐜𝐜𝐜𝐜 × 𝟑𝟑 𝐜𝐜𝐜𝐜��

𝟏𝟏𝟑𝟑 𝐜𝐜𝐜𝐜𝟐𝟐 − 𝟗𝟗.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐

𝟑𝟑.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐

The area is 𝟑𝟑.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐.

There are multiple solution paths for this problem. Explore them with students.

Closing (3 minutes)

What are some helpful methods to use when finding the area of composite areas?

Composing and decomposing the figure into familiar shapes is important. Recording values that are known and marking lengths that are unknown are also very helpful to organize information.

What information and formulas are used in all of the composite area problems?

Usually, the combination of formulas of triangles, rectangles, and circles are used to make up the area of shaded areas. The areas for shaded regions are generally the difference of the area of familiar shapes. Other figures are the sum of the areas of familiar shapes.


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Name Date


Exit Ticket The unshaded regions are quarter circles. Approximate the area of the shaded region. Use 𝜋𝜋 ≈ 3.14.

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Exit Ticket Sample Solutions The unshaded regions are quarter circles. Approximate the area of the shaded region. Use 𝝅𝝅 ≈ 𝟑𝟑.𝟏𝟏𝟏𝟏.

Area of the square − area of the 𝟏𝟏 quarter circles = area of the shaded region

(𝟐𝟐𝟐𝟐 𝐜𝐜 ∙ 𝟐𝟐𝟐𝟐 𝐜𝐜) − ((𝟏𝟏𝟏𝟏 𝐜𝐜)𝟐𝟐 ∙ 𝟑𝟑.𝟏𝟏𝟏𝟏)

𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝟐𝟐 − 𝟑𝟑𝟑𝟑𝟗𝟗.𝟗𝟗𝟏𝟏 𝐜𝐜𝟐𝟐

𝟏𝟏𝟏𝟏𝟏𝟏.𝟏𝟏𝟔𝟔 𝐜𝐜𝟐𝟐

The area of the shaded region is approximately 𝟏𝟏𝟏𝟏𝟏𝟏.𝟏𝟏𝟔𝟔 𝐜𝐜².


1. Find the area of the shaded region. Use 𝟑𝟑.𝟏𝟏𝟏𝟏 for 𝝅𝝅.

Area of large circle– area of small circle

(𝝅𝝅× (𝟐𝟐 𝐜𝐜𝐜𝐜)𝟐𝟐) − (𝝅𝝅× (𝟏𝟏 𝐜𝐜𝐜𝐜)𝟐𝟐)

(𝟑𝟑.𝟏𝟏𝟏𝟏)(𝟔𝟔𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐) − (𝟑𝟑.𝟏𝟏𝟏𝟏)(𝟏𝟏𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐)

𝟐𝟐𝟏𝟏𝟏𝟏.𝟗𝟗𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐 − 𝟓𝟓𝟏𝟏.𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

𝟏𝟏𝟓𝟓𝟏𝟏.𝟑𝟑𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐

The area of the region is approximately 𝟏𝟏𝟓𝟓𝟏𝟏.𝟑𝟑𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐.

2. The figure shows two semicircles. Find the area of the shaded region. Use 𝟑𝟑.𝟏𝟏𝟏𝟏 for 𝝅𝝅.

Area of large semicircle region − area of small semicircle region = area of the shaded region

�𝟏𝟏𝟐𝟐� (𝝅𝝅× (𝟔𝟔 𝐜𝐜𝐜𝐜)𝟐𝟐) − �

𝟏𝟏𝟐𝟐� (𝝅𝝅× (𝟑𝟑 𝐜𝐜𝐜𝐜)𝟐𝟐)

�𝟏𝟏𝟐𝟐� (𝟑𝟑.𝟏𝟏𝟏𝟏)(𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐) − �

𝟏𝟏𝟐𝟐� (𝟑𝟑.𝟏𝟏𝟏𝟏)(𝟗𝟗 𝐜𝐜𝐜𝐜𝟐𝟐)

𝟓𝟓𝟔𝟔.𝟓𝟓𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐 − 𝟏𝟏𝟏𝟏.𝟏𝟏𝟑𝟑 𝐜𝐜𝐜𝐜𝟐𝟐

𝟏𝟏𝟐𝟐.𝟑𝟑𝟗𝟗 𝐜𝐜𝐜𝐜𝟐𝟐

The area is approximately 𝟏𝟏𝟐𝟐.𝟑𝟑𝟗𝟗 𝐜𝐜𝐜𝐜𝟐𝟐.

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3. The figure shows a semicircle and a square. Find the area of the shaded region. Use 𝟑𝟑.𝟏𝟏𝟏𝟏 for 𝝅𝝅.

Area of the square − area of the semicircle

(𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜× 𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜)− �𝟏𝟏𝟐𝟐� ( 𝝅𝝅× (𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜)𝟐𝟐)

𝟓𝟓𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐 − �𝟏𝟏𝟐𝟐� (𝟑𝟑.𝟏𝟏𝟏𝟏× 𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐)

𝟓𝟓𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐 − 𝟐𝟐𝟐𝟐𝟔𝟔.𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐

𝟑𝟑𝟏𝟏𝟗𝟗.𝟗𝟗𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐

The area is approximately 𝟑𝟑𝟏𝟏𝟗𝟗.𝟗𝟗𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐.

4. The figure shows two semicircles and a quarter of a circle. Find the area of the shaded region. Use 𝟑𝟑.𝟏𝟏𝟏𝟏 for 𝝅𝝅.

Area of two semicircles + area of quarter of the larger circle

𝟐𝟐�𝟏𝟏𝟐𝟐� (𝝅𝝅× (𝟓𝟓 𝐜𝐜𝐜𝐜)𝟐𝟐) + �

𝟏𝟏𝟏𝟏� (𝝅𝝅× (𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜)𝟐𝟐)

(𝟑𝟑.𝟏𝟏𝟏𝟏)(𝟐𝟐𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐) + (𝟑𝟑.𝟏𝟏𝟏𝟏)(𝟐𝟐𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐)

𝟑𝟑𝟐𝟐.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟑𝟑𝟐𝟐.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐

𝟏𝟏𝟓𝟓𝟑𝟑 𝐜𝐜𝐜𝐜𝟐𝟐

The area is approximately 𝟏𝟏𝟓𝟓𝟑𝟑 𝐜𝐜𝐜𝐜².

5. Jillian is making a paper flower motif for an art project. The flower she is making has four petals; each petal is formed by three semicircles as shown below. What is the area of the paper flower? Provide your answer in terms of 𝝅𝝅.

Area of medium semicircle + (area of larger semicircle − area of small semicircle)

�𝟏𝟏𝟐𝟐� (𝝅𝝅× (𝟔𝟔 𝐜𝐜𝐜𝐜)𝟐𝟐) + ��

𝟏𝟏𝟐𝟐� (𝝅𝝅 × (𝟗𝟗 𝐜𝐜𝐜𝐜)𝟐𝟐)− �

𝟏𝟏𝟐𝟐� (𝝅𝝅× (𝟑𝟑 𝐜𝐜𝐜𝐜)𝟐𝟐)�

𝟏𝟏𝟐𝟐𝝅𝝅 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟏𝟏𝟏𝟏.𝟓𝟓𝝅𝝅 𝐜𝐜𝐜𝐜𝟐𝟐 − 𝟏𝟏.𝟓𝟓𝝅𝝅 𝐜𝐜𝐜𝐜𝟐𝟐 = 𝟓𝟓𝟏𝟏𝝅𝝅 𝐜𝐜𝐜𝐜𝟐𝟐

𝟓𝟓𝟏𝟏𝝅𝝅 𝐜𝐜𝐜𝐜𝟐𝟐 × 𝟏𝟏

𝟐𝟐𝟏𝟏𝟔𝟔𝝅𝝅 𝐜𝐜𝐜𝐜𝟐𝟐

The area is 𝟐𝟐𝟏𝟏𝟔𝟔𝝅𝝅 𝐜𝐜𝐜𝐜².

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6. The figure is formed by five rectangles. Find the area of the unshaded rectangular region.

Area of the whole rectangle − area of the sum of the shaded rectangles = area of the unshaded rectangular region

(𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜 × 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜) − �𝟐𝟐(𝟑𝟑 𝐜𝐜𝐜𝐜 × 𝟗𝟗 𝐜𝐜𝐜𝐜) + (𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 × 𝟑𝟑 𝐜𝐜𝐜𝐜) + (𝟓𝟓 𝐜𝐜𝐜𝐜 × 𝟗𝟗 𝐜𝐜𝐜𝐜)�

𝟏𝟏𝟔𝟔𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐 − (𝟓𝟓𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟑𝟑𝟑𝟑 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟏𝟏𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐)

𝟏𝟏𝟔𝟔𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐 − 𝟏𝟏𝟑𝟑𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐

𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐

The area is 𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐.

7. The smaller squares in the shaded region each have side lengths of 𝟏𝟏.𝟓𝟓 𝐜𝐜. Find the area of the shaded region.

Area of the 𝟏𝟏𝟔𝟔 𝐜𝐜 by 𝟐𝟐 𝐜𝐜 rectangle − the sum of the area of the smaller unshaded rectangles = area of the shaded region

(𝟏𝟏𝟔𝟔 𝐜𝐜 × 𝟐𝟐 𝐜𝐜)− �(𝟑𝟑 𝐜𝐜× 𝟐𝟐 𝐜𝐜) + �𝟏𝟏(𝟏𝟏.𝟓𝟓 𝐜𝐜× 𝟏𝟏.𝟓𝟓 𝐜𝐜)��

𝟏𝟏𝟐𝟐𝟐𝟐 𝐜𝐜𝟐𝟐 − �𝟔𝟔 𝐜𝐜𝟐𝟐 + 𝟏𝟏(𝟐𝟐.𝟐𝟐𝟓𝟓 𝐜𝐜𝟐𝟐)�

𝟏𝟏𝟐𝟐𝟐𝟐 𝐜𝐜𝟐𝟐 − 𝟏𝟏𝟓𝟓 𝐜𝐜𝟐𝟐

𝟏𝟏𝟏𝟏𝟑𝟑 𝐜𝐜𝟐𝟐

The area is 𝟏𝟏𝟏𝟏𝟑𝟑 𝐜𝐜𝟐𝟐.

8. Find the area of the shaded region.

Area of the sum of the rectangles − area of the right triangle = area of shaded region

�(𝟏𝟏𝟑𝟑 𝐜𝐜𝐜𝐜× 𝟏𝟏 𝐜𝐜𝐜𝐜) + (𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜× 𝟐𝟐 𝐜𝐜𝐜𝐜)� − ��𝟏𝟏𝟐𝟐� (𝟏𝟏𝟑𝟑 𝐜𝐜𝐜𝐜 × 𝟑𝟑 𝐜𝐜𝐜𝐜)�

(𝟔𝟔𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟏𝟏𝟔𝟔𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐) − �𝟏𝟏𝟐𝟐� (𝟗𝟗𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐)

𝟐𝟐𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐 − 𝟏𝟏𝟓𝟓.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐

𝟏𝟏𝟗𝟗𝟏𝟏.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐

The area is 𝟏𝟏𝟗𝟗𝟏𝟏.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐.

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9.

a. Find the area of the shaded region.

Area of the two parallelograms − area of square in the center = area of the shaded region

𝟐𝟐(𝟓𝟓 𝐜𝐜𝐜𝐜× 𝟏𝟏𝟔𝟔 𝐜𝐜𝐜𝐜)− (𝟏𝟏 𝐜𝐜𝐜𝐜 × 𝟏𝟏 𝐜𝐜𝐜𝐜)

𝟏𝟏𝟔𝟔𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 − 𝟏𝟏𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐

𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

The area is 𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐.

b. Draw two ways the figure above can be divided in four equal parts.

c. What is the area of one of the parts in (b)?

𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 ÷ 𝟏𝟏 = 𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐

The area of one of the parts in (b) is 𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐.

10. The figure is a rectangle made out of triangles. Find the area of the shaded region.

Area of the rectangle − area of the unshaded triangles = area of the shaded region

(𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜 × 𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜)− ��𝟏𝟏𝟐𝟐� (𝟗𝟗 𝐜𝐜𝐜𝐜 × 𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜) + �

𝟏𝟏𝟐𝟐� (𝟗𝟗 𝐜𝐜𝐜𝐜 × 𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜)�

𝟓𝟓𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 − (𝟗𝟗𝟏𝟏.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐)

𝟓𝟓𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 − 𝟐𝟐𝟏𝟏𝟐𝟐.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐

𝟑𝟑𝟏𝟏𝟏𝟏.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐

The area is 𝟑𝟑𝟏𝟏𝟏𝟏.𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐.

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11. The figure consists of a right triangle and an eighth of a circle. Find the area of the shaded region. Use 𝟐𝟐𝟐𝟐𝟑𝟑

for 𝝅𝝅.

Area of right triangle – area of eighth of the circle= area of shaded region

�𝟏𝟏𝟐𝟐� (𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜× 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜)− �

𝟏𝟏𝟐𝟐� (𝛑𝛑× 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 × 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜)

�𝟏𝟏𝟐𝟐� (𝟏𝟏𝟗𝟗𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐) − �

𝟏𝟏𝟐𝟐� �𝟐𝟐𝟐𝟐𝟑𝟑� (𝟐𝟐 𝐜𝐜𝐜𝐜 × 𝟑𝟑 𝐜𝐜𝐜𝐜 × 𝟐𝟐 𝐜𝐜𝐜𝐜 × 𝟑𝟑 𝐜𝐜𝐜𝐜)

𝟗𝟗𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐 − 𝟑𝟑𝟑𝟑 𝐜𝐜𝐜𝐜𝟐𝟐

𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

The area is approximately 𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐.

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7•3 Lesson 21

Lesson 21: Surface Area

𝟒𝟒 𝟑𝟑 𝟔𝟔 𝟔𝟔 𝟑𝟑 𝟔𝟔 𝟒𝟒 𝟑𝟑 𝟔𝟔


Student Outcomes

Students find the surface area of three-dimensional objects whose surface area is composed of triangles and quadrilaterals. They use polyhedron nets to understand that surface area is simply the sum of the area of the lateral faces and the area of the base(s).

Classwork

Opening Exercise (8 minutes): Surface Area of a Right Rectangular Prism

Students use prior knowledge to find the surface area of the given right rectangular prism by decomposing the prism into the plane figures that represent its individual faces. Students then discuss their methods aloud.

Opening Exercise: Surface Area of a Right Rectangular Prism

On the provided grid, draw a net representing the surfaces of the right rectangular prism (assume each grid line represents 𝟏𝟏 inch). Then, find the surface area of the prism by finding the area of the net.

There are six rectangular faces that make up the net.

The four rectangles in the center form one long rectangle that is 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢. by 𝟑𝟑 𝐢𝐢𝐢𝐢.

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝒍𝒍𝒍𝒍

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟑𝟑 𝐢𝐢𝐢𝐢 ∙ 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟔𝟔𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐

Two rectangles form the wings, both 𝟔𝟔 𝐢𝐢𝐢𝐢 by 𝟒𝟒 𝐢𝐢𝐢𝐢.

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝒍𝒍𝒍𝒍

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟔𝟔 𝐢𝐢𝐢𝐢 ∙ 𝟒𝟒 𝐢𝐢𝐢𝐢

𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 = 𝟐𝟐𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐

The area of both wings is 𝟐𝟐(𝟐𝟐𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐) = 𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐.

The total area of the net is

𝑨𝑨 = 𝟔𝟔𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐

The net represents all the surfaces of the rectangular prism, so its area is equal to the surface area of the prism. The surface area of the right rectangular prism is 𝟏𝟏𝟐𝟐𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐.

Note: Students may draw any of the variations of nets for the given prism.

Scaffolding: Students may need to review the meaning of the term net from Grade 6. Prepare a solid right rectangular prism such as a wooden block and a paper net covering the prism to model where a net comes from.

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3 in

6 in

4 in

5 in


What other ways could we have found the surface area of the rectangular prism?

Surface area formula: 𝑆𝑆𝑆𝑆 = 2𝑙𝑙𝑙𝑙 + 2𝑙𝑙ℎ + 2𝑙𝑙ℎ 𝑆𝑆𝑆𝑆 = 2(3 in ∙ 4 in) + 2(3 in ∙ 6 in) + 2(4 in ∙ 6 in) 𝑆𝑆𝑆𝑆 = 24 in2 + 36 in2 + 48 in2 𝑆𝑆𝑆𝑆 = 108 in2

Find the areas of each individual rectangular face:

Area = length × width 𝑆𝑆 = 6 in × 3 in 𝑆𝑆 = 4 in × 3 in 𝑆𝑆 = 6 in × 4 in 𝑆𝑆 = 18 in2 𝑆𝑆 = 12 in2 𝑆𝑆 = 24 in2 There are two of each face, so 𝑆𝑆𝑆𝑆 = 2(18 in2 + 12 in2 + 24 in2) 𝑆𝑆𝑆𝑆 = 2(54 in2) 𝑆𝑆𝑆𝑆 = 108 in2.

Discussion (6 minutes): Terminology

A right prism can be described as a solid with two “end” faces (called its bases) that are exact copies of each other and rectangular faces that join corresponding edges of the bases (called lateral faces).

Are the bottom and top faces of a right rectangular prism the bases of the prism?

Not always. Any of its opposite faces can be considered bases because they are all rectangles.

If we slice the right rectangular prism in half along a diagonal of a base (see picture), the two halves are called right triangular prisms. Why do you think they are called triangular prisms?

The bases of each prism are triangles, and prisms are named by their bases.

Why must the triangular faces be the bases of these prisms?

Because the lateral faces (faces that are not bases) of a right prism have to be rectangles.

6 in 6 in

3 in

3 in

4 in

4 in

Front rectangle

Top rectangle

Side rectangle

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Can the surface area formula for a right rectangular prism (𝑆𝑆𝑆𝑆 = 2𝑙𝑙𝑙𝑙 + 2𝑙𝑙ℎ + 2𝑙𝑙ℎ) be applied to find the surface area of a right triangular prism? Why or why not?

No, because each of the terms in the surface area formula represents the area of a rectangular face. A right triangular prism has bases that are triangular, not rectangular.


Students find the surface area of the right triangular prism to determine the validity of a given conjecture.

Exercise 1

Marcus thinks that the surface area of the right triangular prism will be half that of the right rectangular prism and wants

to use the modified formula 𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟐𝟐 (𝟐𝟐𝒍𝒍𝒍𝒍+ 𝟐𝟐𝒍𝒍𝟐𝟐 + 𝟐𝟐𝒍𝒍𝟐𝟐). Do you agree or disagree with Marcus? Use nets of the prisms

to support your argument.

The surface area of the right rectangular prism is 𝟏𝟏𝟐𝟐𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐, so Marcus believes the surface areas of each right triangular prism is 𝟓𝟓𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐.

Students can make comparisons of the area values depicted in the nets of the prisms and can also compare the physical areas of the nets either by overlapping the nets on the same grid or using a transparent overlay.

The net of the right triangular prism has one less face than the right rectangular prism. Two of the rectangular faces on the right triangular prism (rectangular regions 1 and 2 in the diagram) are the same faces from the right rectangular prism, so they are the same size. The areas of the triangular bases (triangular regions 3 and 4 in the diagram) are half the area of their corresponding rectangular faces of the right rectangular prism. These four faces of the right triangular prism make up half the surface area of the right rectangular prism before considering the fifth face; no, Marcus is incorrect.

The areas of rectangular faces 1 and 2, plus the areas of the triangular regions 3 and 4 is 𝟓𝟓𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐. The last rectangular region has an area of 𝟑𝟑𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐. The total area of the net is 𝟓𝟓𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟑𝟑𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 or 𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐, which is far more than half the surface area of the right rectangular prism.

Use a transparency to show students how the nets overlap where the lateral faces together form a longer rectangular region, and the bases are represented by “wings” on either side of that triangle. Consider using student work for this if there is a good example. Use this setup in the following discussion.


The surface area formula (𝑆𝑆𝑆𝑆 = 2𝑙𝑙𝑙𝑙 + 2𝑙𝑙ℎ + 2𝑙𝑙ℎ) for a right rectangular prism cannot be applied to a right triangular prism. Why?

The formula adds the areas of six rectangular faces. A right triangular prism only has three rectangular faces and also has two triangular faces (bases).

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The area formula for triangles is 12

the formula for the area of rectangles or parallelograms. Can the surface

area of a triangular prism be obtained by dividing the surface area formula for a right rectangular prism by 2? Explain.

No. The right triangular prism in the above example had more than half the surface area of the right rectangular prism that it was cut from. If this occurs in one case, then it must occur in others as well.

If you compare the nets of the right rectangular prism and the right triangular prism, what do the nets seem to have in common? (Hint: What do all right prisms have in common? Answer: Rectangular lateral faces) Their lateral faces form a larger rectangular region, and the bases are

attached to the side of that rectangle like “wings.”

Will this commonality always exist in right prisms? How do you know?

Yes. Right prisms must have rectangular lateral faces. If we align all the lateral faces of a right prism in a net, they can always form a larger rectangular region because they all have the same height as the prism.

How do we determine the total surface area of the prism?

Add the total area of the lateral faces and the areas of the bases.

If we let 𝐿𝐿𝑆𝑆 represent the lateral area and let 𝐵𝐵 represent the area of a base, then the surface area of a right prism can be found using the formula:

𝑆𝑆𝑆𝑆 = 𝐿𝐿𝑆𝑆 + 2𝐵𝐵.

Example 1 (6 minutes): Lateral Area of a Right Prism

Students find the lateral areas of right prisms and recognize the pattern of multiplying the height of the right prism (the distance between its bases) by the perimeter of the prism’s base.

Example 1: Lateral Area of a Right Prism

A right triangular prism, a right rectangular prism, and a right pentagonal prism are pictured below, and all have equal heights of 𝟐𝟐.

a. Write an expression that represents the lateral area of the right triangular prism as the sum of the areas of its lateral faces.

𝒂𝒂 ∙ 𝟐𝟐 + 𝒃𝒃 ∙ 𝟐𝟐 + 𝒄𝒄 ∙ 𝟐𝟐

b. Write an expression that represents the lateral area of the right rectangular prism as the sum of the areas of its lateral faces.

𝒂𝒂 ∙ 𝟐𝟐 + 𝒃𝒃 ∙ 𝟐𝟐 + 𝒂𝒂 ∙ 𝟐𝟐+ 𝒃𝒃 ∙ 𝟐𝟐

Scaffolding: The teacher may need to assist students in finding the commonality between the nets of right prisms by showing examples of various right prisms and pointing out the fact that they all have rectangular lateral faces. The rectangular faces may be described as “connectors” between the bases of a right prism.

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c. Write an expression that represents the lateral area of the right pentagonal prism as the sum of the areas of its lateral faces.

𝒂𝒂 ∙ 𝟐𝟐 + 𝒃𝒃 ∙ 𝟐𝟐 + 𝒄𝒄 ∙ 𝟐𝟐 + 𝒅𝒅 ∙ 𝟐𝟐 + 𝒆𝒆 ∙ 𝟐𝟐

d. What value appears often in each expression and why?

𝟐𝟐; Each prism has a height of 𝟐𝟐; therefore, each lateral face has a height of h.

e. Rewrite each expression in factored form using the distributive property and the height of each lateral face.

𝟐𝟐(𝒂𝒂 + 𝒃𝒃 + 𝒄𝒄) 𝟐𝟐(𝒂𝒂 + 𝒃𝒃 + 𝒂𝒂 + 𝒃𝒃) 𝟐𝟐(𝒂𝒂 + 𝒃𝒃 + 𝒄𝒄 + 𝒅𝒅+ 𝒆𝒆)

f. What do the parentheses in each case represent with respect to the right prisms?

𝟐𝟐 (𝒂𝒂 + 𝒃𝒃 + 𝒄𝒄)��𝐩𝐩𝐀𝐀𝐀𝐀𝐢𝐢𝐩𝐩𝐀𝐀𝐩𝐩𝐀𝐀𝐀𝐀

𝟐𝟐 (𝒂𝒂 + 𝒃𝒃 + 𝒂𝒂 + 𝒃𝒃)��𝐩𝐩𝐀𝐀𝐀𝐀𝐢𝐢𝐩𝐩𝐀𝐀𝐩𝐩𝐀𝐀𝐀𝐀

𝟐𝟐 (𝒂𝒂 + 𝒃𝒃 + 𝒄𝒄 + 𝒅𝒅+ 𝒆𝒆)��𝐩𝐩𝐀𝐀𝐀𝐀𝐢𝐢𝐩𝐩𝐀𝐀𝐩𝐩𝐀𝐀𝐀𝐀

The perimeter of the base of the corresponding prism.

g. How can we generalize the lateral area of a right prism into a formula that applies to all right prisms?

If 𝑳𝑳𝑨𝑨 represents the lateral area of a right prism, 𝑷𝑷 represents the perimeter of the right prism’s base, and 𝟐𝟐 represents the distance between the right prism’s bases, then:

𝑳𝑳𝑨𝑨 = 𝑷𝑷𝐛𝐛𝐀𝐀𝐛𝐛𝐀𝐀 ∙ 𝟐𝟐.

Closing (5 minutes)

The vocabulary below contains the precise definitions of the visual and colloquial descriptions used in the lesson. Please read through the definitions aloud with your students, asking questions that compare the visual and colloquial descriptions used in the lesson with the precise definitions.

Relevant Vocabulary

RIGHT PRISM: Let 𝑬𝑬 and 𝑬𝑬′ be two parallel planes. Let 𝑩𝑩 be a triangular or rectangular region or a region that is the union of such regions in the plane 𝑬𝑬. At each point 𝑷𝑷 of 𝑩𝑩, consider the segment 𝑷𝑷𝑷𝑷′ perpendicular to 𝑬𝑬, joining 𝑷𝑷 to a point 𝑷𝑷′ of the plane 𝑬𝑬′. The union of all these segments is a solid called a right prism.

There is a region 𝑩𝑩′ in 𝑬𝑬′ that is an exact copy of the region 𝑩𝑩. The regions 𝑩𝑩 and 𝑩𝑩′ are called the base faces (or just bases) of the prism. The rectangular regions between two corresponding sides of the bases are called lateral faces of the prism. In all, the boundary of a right rectangular prism has 𝟔𝟔 faces: 𝟐𝟐 base faces and 𝟒𝟒 lateral faces. All adjacent faces intersect along segments called edges (base edges and lateral edges).

Scaffolding: Example 1 can be explored further by assigning numbers to represent the lengths of the sides of the bases of each prism. If students represent the lateral area as the sum of the areas of the lateral faces without evaluating, the common factor in each term will be evident and can then be factored out to reveal the same relationship.

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CUBE: A cube is a right rectangular prism all of whose edges are of equal length.

SURFACE: The surface of a prism is the union of all of its faces (the base faces and lateral faces).

NET: A net is a two-dimensional diagram of the surface of a prism.

1. Why are the lateral faces of right prisms always rectangular regions?

Because along a base edge, the line segments 𝑷𝑷𝑷𝑷′ are always perpendicular to the edge, forming a rectangular region.

2. What is the name of the right prism whose bases are rectangles?

Right rectangular prism

3. How does this definition of right prism include the interior of the prism?

The union of all the line segments fills out the interior.


Lesson Summary

The surface area of a right prism can be obtained by adding the areas of the lateral faces to the area of the bases. The formula for the surface area of a right prism is 𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟐𝟐𝑩𝑩, where 𝑺𝑺𝑨𝑨 represents the surface area of the prism, 𝑳𝑳𝑨𝑨 represents the area of the lateral faces, and 𝑩𝑩 represents the area of one base. The lateral area 𝑳𝑳𝑨𝑨 can be obtained by multiplying the perimeter of the base of the prism times the height of the prism.

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7•3 Lesson 21


Name Date


Exit Ticket Find the surface area of the right trapezoidal prism. Show all necessary work.

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Exit Ticket Sample Solutions Find the surface area of the right trapezoidal prism. Show all necessary work.

𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟐𝟐𝑩𝑩

𝑳𝑳𝑨𝑨 = 𝑷𝑷 ∙ 𝟐𝟐

𝑳𝑳𝑨𝑨 = (𝟑𝟑 𝐜𝐜𝐩𝐩 + 𝟕𝟕 𝐜𝐜𝐩𝐩 + 𝟓𝟓 𝐜𝐜𝐩𝐩 + 𝟏𝟏𝟏𝟏 𝐜𝐜𝐩𝐩 ) ∙ 𝟔𝟔 𝐜𝐜𝐩𝐩

𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟔𝟔 𝐜𝐜𝐩𝐩 ∙ 𝟔𝟔 𝐜𝐜𝐩𝐩

𝑳𝑳𝑨𝑨 = 𝟏𝟏𝟓𝟓𝟔𝟔 𝐜𝐜𝐩𝐩𝟐𝟐

Each base consists of a 𝟑𝟑 𝐜𝐜𝐩𝐩 by 𝟕𝟕 𝐜𝐜𝐩𝐩 rectangle and right triangle with a base of 𝟑𝟑 𝐜𝐜𝐩𝐩 and a height of 𝟒𝟒 𝐜𝐜𝐩𝐩. Therefore, the area of each base:

𝑩𝑩 = 𝑨𝑨𝒓𝒓 + 𝑨𝑨𝒕𝒕

𝑩𝑩 = 𝒍𝒍𝒍𝒍 + 𝟏𝟏𝟐𝟐𝒃𝒃𝟐𝟐

𝑩𝑩 = (𝟕𝟕 𝐜𝐜𝐩𝐩 ∙ 𝟑𝟑 𝐜𝐜𝐩𝐩) + �𝟏𝟏𝟐𝟐 ∙ 𝟑𝟑 𝐜𝐜𝐩𝐩 ∙ 𝟒𝟒 𝐜𝐜𝐩𝐩�

𝑩𝑩 = 𝟐𝟐𝟏𝟏 𝐜𝐜𝐩𝐩𝟐𝟐 + 𝟔𝟔 𝐜𝐜𝐩𝐩𝟐𝟐

𝑩𝑩 = 𝟐𝟐𝟕𝟕 𝐜𝐜𝐩𝐩𝟐𝟐


𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟓𝟓𝟔𝟔 𝐜𝐜𝐩𝐩𝟐𝟐 + 𝟐𝟐(𝟐𝟐𝟕𝟕 𝐜𝐜𝐩𝐩𝟐𝟐)

𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟓𝟓𝟔𝟔 𝐜𝐜𝐩𝐩𝟐𝟐 + 𝟓𝟓𝟒𝟒 𝐜𝐜𝐩𝐩𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐

The surface of the right trapezoidal prism is 𝟐𝟐𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐.


1. For each of the following nets, highlight the perimeter of the lateral area, draw the solid represented by the net, indicate the type of solid, and then find the solid’s surface area.

a. Right rectangular prism


𝑳𝑳𝑨𝑨 = 𝑷𝑷 ∙ 𝟐𝟐 𝑩𝑩 = 𝒍𝒍𝒍𝒍

𝑳𝑳𝑨𝑨 = �𝟐𝟐𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩 + 𝟕𝟕𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩 + 𝟐𝟐𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩 + 𝟕𝟕𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩� ∙ 𝟓𝟓 𝐜𝐜𝐩𝐩 𝑩𝑩 = 𝟐𝟐𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩 ∙ 𝟕𝟕𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩

𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟐𝟐 𝐜𝐜𝐩𝐩 ∙ 𝟓𝟓 𝐜𝐜𝐩𝐩 𝑩𝑩 = 𝟓𝟓𝟐𝟐 𝐜𝐜𝐩𝐩 ∙ 𝟏𝟏𝟓𝟓𝟐𝟐 𝐜𝐜𝐩𝐩

𝑳𝑳𝑨𝑨 = 𝟏𝟏𝟐𝟐𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 𝑩𝑩 = 𝟕𝟕𝟓𝟓𝟒𝟒 𝐜𝐜𝐩𝐩𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟐𝟐𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 + 𝟐𝟐�𝟕𝟕𝟓𝟓𝟒𝟒 𝐜𝐜𝐩𝐩𝟐𝟐�

𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟐𝟐𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 + 𝟑𝟑𝟕𝟕.𝟓𝟓 𝐜𝐜𝐩𝐩𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟑𝟑𝟕𝟕.𝟓𝟓 𝐜𝐜𝐩𝐩𝟐𝟐

The surface area of the right rectangular prism is 𝟏𝟏𝟑𝟑𝟕𝟕.𝟓𝟓 𝐜𝐜𝐩𝐩𝟐𝟐

(3-Dimensional Form)

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7•3 Lesson 21


b. Right triangular prism


𝑳𝑳𝑨𝑨 = 𝑷𝑷 ∙ 𝟐𝟐 𝑩𝑩 = 𝟏𝟏𝟐𝟐𝒃𝒃𝟐𝟐

𝑳𝑳𝑨𝑨 = (𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. + 𝟒𝟒 𝐢𝐢𝐢𝐢. + 𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. ) ∙ 𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. 𝑩𝑩 = 𝟏𝟏𝟐𝟐 (𝟒𝟒 𝐢𝐢𝐢𝐢. )�𝟗𝟗𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢. �

𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟒𝟒 𝐢𝐢𝐢𝐢.∙ 𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. 𝑩𝑩 = 𝟒𝟒 𝐢𝐢𝐢𝐢. �𝟗𝟗𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢. �

𝑳𝑳𝑨𝑨 = 𝟑𝟑𝟑𝟑𝟔𝟔 𝐢𝐢𝐢𝐢𝟐𝟐 𝑩𝑩 = �𝟑𝟑𝟔𝟔+ 𝟒𝟒𝟓𝟓� 𝐢𝐢𝐢𝐢

𝟐𝟐

𝑩𝑩 = 𝟑𝟑𝟔𝟔𝟒𝟒𝟓𝟓 𝐢𝐢𝐢𝐢𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟑𝟑𝟔𝟔 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟐𝟐�𝟑𝟑𝟔𝟔𝟒𝟒𝟓𝟓 𝐢𝐢𝐢𝐢𝟐𝟐�

𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟑𝟑𝟔𝟔 𝐢𝐢𝐢𝐢𝟐𝟐 + �𝟕𝟕𝟐𝟐+ 𝟒𝟒𝟓𝟓� 𝐢𝐢𝐢𝐢

𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟐𝟐𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟏𝟏𝟑𝟑𝟓𝟓 𝐢𝐢𝐢𝐢𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟐𝟐𝟗𝟗𝟑𝟑𝟓𝟓 𝐢𝐢𝐢𝐢𝟐𝟐

The surface area of the right triangular prism is 𝟒𝟒𝟐𝟐𝟗𝟗𝟑𝟑𝟓𝟓 𝐢𝐢𝐢𝐢𝟐𝟐.

2. Given a cube with edges that are 𝟑𝟑𝟒𝟒 inch long:

a. Find the surface area of the cube.

𝑺𝑺𝑨𝑨 = 𝟔𝟔𝒔𝒔𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟔𝟔�𝟑𝟑𝟒𝟒

𝐢𝐢𝐢𝐢. �𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟔𝟔�𝟑𝟑𝟒𝟒

𝐢𝐢𝐢𝐢. � ∙ �𝟑𝟑𝟒𝟒

𝐢𝐢𝐢𝐢. �

𝑺𝑺𝑨𝑨 = 𝟔𝟔�𝟗𝟗𝟏𝟏𝟔𝟔

𝐢𝐢𝐢𝐢𝟐𝟐�

𝑺𝑺𝑨𝑨 =𝟐𝟐𝟕𝟕𝟒𝟒

𝐢𝐢𝐢𝐢𝟐𝟐 or 𝟑𝟑𝟑𝟑𝟒𝟒

𝐢𝐢𝐢𝐢𝟐𝟐

b. Joshua makes a scale drawing of the cube using a scale factor of 𝟒𝟒. Find the surface area of the cube that Joshua drew.

𝟑𝟑𝟒𝟒𝐢𝐢𝐢𝐢.∙ 𝟒𝟒 = 𝟑𝟑 𝐢𝐢𝐢𝐢.; The edge lengths of Joshua’s drawing would be 𝟑𝟑 inches.

𝑺𝑺𝑨𝑨 = 𝟔𝟔(𝟑𝟑 𝐢𝐢𝐢𝐢. )𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟔𝟔(𝟗𝟗 𝐢𝐢𝐢𝐢𝟐𝟐) 𝑺𝑺𝑨𝑨 = 𝟓𝟓𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐


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7•3 Lesson 21


c. What is the ratio of the surface area of the scale drawing to the surface area of the actual cube, and how does the value of the ratio compare to the scale factor?

𝟓𝟓𝟒𝟒÷ 𝟑𝟑𝟑𝟑𝟒𝟒

𝟓𝟓𝟒𝟒÷ 𝟐𝟐𝟕𝟕𝟒𝟒

𝟓𝟓𝟒𝟒 ∙ 𝟒𝟒𝟐𝟐𝟕𝟕

𝟐𝟐 ∙ 𝟒𝟒 = 𝟏𝟏𝟔𝟔. The ratios of the surface area of the scale drawing to the surface area of the actual cube is 𝟏𝟏𝟔𝟔:𝟏𝟏. The value of the ratio is 𝟏𝟏𝟔𝟔. The scale factor of the drawing is 𝟒𝟒, and the value of the ratio of the surface area of the drawing to the surface area of the actual cube is 𝟒𝟒𝟐𝟐 or 𝟏𝟏𝟔𝟔.

3. Find the surface area of each of the following right prisms using the formula 𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨 + 𝟐𝟐𝑩𝑩.

a.

𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨 + 𝟐𝟐𝑩𝑩


𝑳𝑳𝑨𝑨 = �𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐

𝐩𝐩𝐩𝐩+ 𝟏𝟏𝟐𝟐 𝐩𝐩𝐩𝐩 + 𝟕𝟕𝟏𝟏𝟐𝟐

𝐩𝐩𝐩𝐩� ∙ 𝟏𝟏𝟓𝟓 𝐩𝐩𝐩𝐩

𝑳𝑳𝑨𝑨 = 𝟑𝟑𝟐𝟐 𝐩𝐩𝐩𝐩 ∙ 𝟏𝟏𝟓𝟓 𝐩𝐩𝐩𝐩

𝑳𝑳𝑨𝑨 = 𝟒𝟒𝟓𝟓𝟐𝟐 𝐩𝐩𝐩𝐩𝟐𝟐

𝑩𝑩 = 𝟏𝟏𝟐𝟐𝒃𝒃𝟐𝟐 𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟓𝟓𝟐𝟐 𝐩𝐩𝐩𝐩𝟐𝟐 + 𝟐𝟐�𝟕𝟕𝟓𝟓𝟐𝟐 𝐩𝐩𝐩𝐩𝟐𝟐�

𝑩𝑩 = 𝟏𝟏𝟐𝟐 ∙ �𝟕𝟕

𝟏𝟏𝟐𝟐 𝐩𝐩𝐩𝐩� ∙ (𝟏𝟏𝟐𝟐 𝐩𝐩𝐩𝐩) 𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟓𝟓𝟐𝟐 𝐩𝐩𝐩𝐩𝟐𝟐 + 𝟕𝟕𝟓𝟓 𝐩𝐩𝐩𝐩𝟐𝟐

𝑩𝑩 = 𝟏𝟏𝟐𝟐 ∙ (𝟕𝟕𝟐𝟐+ 𝟓𝟓) 𝐩𝐩𝐩𝐩𝟐𝟐 𝑺𝑺𝑨𝑨 = 𝟓𝟓𝟐𝟐𝟓𝟓 𝐩𝐩𝐩𝐩𝟐𝟐

𝑩𝑩 = 𝟏𝟏𝟐𝟐 ∙ 𝟕𝟕𝟓𝟓 𝐩𝐩𝐩𝐩𝟐𝟐

𝑩𝑩 = 𝟕𝟕𝟓𝟓𝟐𝟐 𝐩𝐩𝐩𝐩𝟐𝟐

The surface area of the prism is 𝟓𝟓𝟐𝟐𝟓𝟓 𝐩𝐩𝐩𝐩𝟐𝟐.

𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐 mm

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7•3 Lesson 21


b.


𝑳𝑳𝑨𝑨 = 𝑷𝑷 ∙ 𝟐𝟐 𝑩𝑩 = 𝟏𝟏𝟐𝟐𝒃𝒃𝟐𝟐

𝑳𝑳𝑨𝑨 = �𝟗𝟗 𝟑𝟑𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢. +𝟔𝟔𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. +𝟒𝟒 𝐢𝐢𝐢𝐢. � ∙ 𝟓𝟓 𝐢𝐢𝐢𝐢 𝑩𝑩 = 𝟏𝟏

𝟐𝟐 ∙ 𝟗𝟗𝟑𝟑𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢.∙ 𝟐𝟐 𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢.

𝑳𝑳𝑨𝑨 = �𝟐𝟐𝟐𝟐𝟒𝟒𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢. +𝟏𝟏𝟑𝟑𝟐𝟐 𝐢𝐢𝐢𝐢. +𝟒𝟒 𝐢𝐢𝐢𝐢. � ∙ 𝟓𝟓 𝐢𝐢𝐢𝐢 𝑩𝑩 = 𝟏𝟏

𝟐𝟐 ∙𝟐𝟐𝟐𝟐𝟒𝟒𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢.∙ 𝟓𝟓𝟐𝟐 𝐢𝐢𝐢𝐢.

𝑳𝑳𝑨𝑨 = �𝟒𝟒𝟓𝟓𝟔𝟔𝟓𝟓𝟐𝟐 𝐢𝐢𝐢𝐢. +𝟑𝟑𝟐𝟐𝟓𝟓𝟓𝟓𝟐𝟐 𝐢𝐢𝐢𝐢. +𝟐𝟐𝟐𝟐𝟐𝟐

𝟓𝟓𝟐𝟐 𝐢𝐢𝐢𝐢. � ∙ 𝟓𝟓 𝐢𝐢𝐢𝐢. 𝑩𝑩 = 𝟏𝟏,𝟏𝟏𝟒𝟒𝟐𝟐𝟏𝟏𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢2

𝑳𝑳𝑨𝑨 = �𝟗𝟗𝟒𝟒𝟏𝟏𝟓𝟓𝟐𝟐 𝐢𝐢𝐢𝐢. � ∙ 𝟓𝟓 𝐢𝐢𝐢𝐢. 𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢2

𝑳𝑳𝑨𝑨 = 𝟒𝟒𝟗𝟗,𝟐𝟐𝟓𝟓𝟐𝟐𝟓𝟓𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 𝟐𝟐𝑩𝑩 = 𝟐𝟐 ∙ 𝟏𝟏𝟏𝟏𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢

𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟗𝟗𝟒𝟒 𝟏𝟏𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 𝟐𝟐𝑩𝑩 = 𝟐𝟐𝟐𝟐𝟒𝟒𝟓𝟓 𝐢𝐢𝐢𝐢

𝟐𝟐


𝑺𝑺𝑨𝑨 = 𝟗𝟗𝟒𝟒 𝟏𝟏𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟒𝟒𝟓𝟓 𝐢𝐢𝐢𝐢𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟐𝟐𝟐𝟐 𝟗𝟗𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐

The surface area of the prism is 𝟏𝟏𝟐𝟐𝟐𝟐 𝟗𝟗𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐.

c.



𝑳𝑳𝑨𝑨 = �𝟏𝟏𝟒𝟒

𝐢𝐢𝐢𝐢. +𝟏𝟏𝟐𝟐

𝐢𝐢𝐢𝐢. +𝟏𝟏𝟒𝟒𝐢𝐢𝐢𝐢. +

𝟏𝟏𝟒𝟒

𝐢𝐢𝐢𝐢. +𝟏𝟏𝟐𝟐𝐢𝐢𝐢𝐢. +

𝟏𝟏𝟒𝟒

𝐢𝐢𝐢𝐢. � ∙ 𝟐𝟐 𝐢𝐢𝐢𝐢.

𝑳𝑳𝑨𝑨 = �𝟏𝟏𝟑𝟑𝟒𝟒 𝐢𝐢𝐢𝐢. � ∙ 𝟐𝟐 𝐢𝐢𝐢𝐢.

𝑳𝑳𝑨𝑨 = 𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 𝑩𝑩 = 𝑨𝑨𝒓𝒓𝒆𝒆𝒄𝒄𝒕𝒕𝒂𝒂𝒓𝒓𝒓𝒓𝒍𝒍𝒆𝒆 + 𝟐𝟐𝑨𝑨𝒕𝒕𝒓𝒓𝒕𝒕𝒂𝒂𝒓𝒓𝒓𝒓𝒍𝒍𝒆𝒆

𝑳𝑳𝑨𝑨 = 𝟑𝟑𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 𝑩𝑩 = �𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢.∙ 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢. � + 𝟐𝟐 ∙ 𝟏𝟏𝟐𝟐 �𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢.∙ 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢. �

𝑩𝑩 = � 𝟏𝟏𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐�+ � 𝟏𝟏𝟒𝟒𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐�

𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟏𝟏𝟐𝟐 𝒕𝒕𝒓𝒓𝟐𝟐 + 𝟐𝟐�𝟏𝟏𝟒𝟒 𝒕𝒕𝒓𝒓𝟐𝟐� 𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟏𝟏

𝟒𝟒𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟏𝟏𝟐𝟐 𝒕𝒕𝒓𝒓𝟐𝟐 + 𝟏𝟏𝟒𝟒 𝒕𝒕𝒓𝒓𝟐𝟐 𝑩𝑩 = 𝟒𝟒

𝟒𝟒𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟏𝟏𝟒𝟒𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟐𝟐𝟒𝟒 𝒕𝒕𝒓𝒓𝟐𝟐 + 𝟏𝟏𝟒𝟒 𝒕𝒕𝒓𝒓𝟐𝟐 𝑩𝑩 = 𝟓𝟓

𝟒𝟒𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟑𝟑𝟒𝟒 𝒕𝒕𝒓𝒓𝟐𝟐 𝑩𝑩 = 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐

The surface area of the prism is 𝟑𝟑𝟑𝟑𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐.

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d.



𝑳𝑳𝑨𝑨 = (𝟏𝟏𝟑𝟑 𝐜𝐜𝐩𝐩+ 𝟏𝟏𝟑𝟑 𝐜𝐜𝐩𝐩 + 𝟒𝟒.𝟔𝟔 𝐜𝐜𝐩𝐩 + 𝟒𝟒.𝟔𝟔 𝐜𝐜𝐩𝐩) ∙ 𝟐𝟐𝟏𝟏𝟒𝟒𝐜𝐜𝐩𝐩

𝑳𝑳𝑨𝑨 = (𝟐𝟐𝟔𝟔 𝐜𝐜𝐩𝐩+ 𝟏𝟏𝟕𝟕.𝟐𝟐 𝐜𝐜𝐩𝐩) ∙ 𝟐𝟐𝟏𝟏𝟒𝟒

𝐜𝐜𝐩𝐩

𝑳𝑳𝑨𝑨 = (𝟒𝟒𝟑𝟑.𝟐𝟐)𝐜𝐜𝐩𝐩 ∙ 𝟐𝟐𝟏𝟏𝟒𝟒 𝐜𝐜𝐩𝐩

𝑳𝑳𝑨𝑨 = (𝟒𝟒𝟔𝟔.𝟒𝟒 𝐜𝐜𝐩𝐩𝟐𝟐 + 𝟏𝟏𝟐𝟐.𝟒𝟒 𝐜𝐜𝐩𝐩𝟐𝟐)

𝑳𝑳𝑨𝑨 = 𝟗𝟗𝟕𝟕.𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨 + 𝟐𝟐𝑩𝑩 𝑩𝑩 = 𝟏𝟏𝟐𝟐 (𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩 ∙ 𝟕𝟕 𝐜𝐜𝐩𝐩 ) + 𝟏𝟏

𝟐𝟐 (𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩 ∙ 𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩 )

𝑺𝑺𝑨𝑨 = 𝟗𝟗𝟕𝟕.𝟐𝟐 𝒄𝒄𝒎𝒎𝟐𝟐 + 𝟐𝟐(𝟗𝟗𝟓𝟓 𝐜𝐜𝐩𝐩𝟐𝟐) 𝑩𝑩 = 𝟏𝟏𝟐𝟐 (𝟕𝟕𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 + 𝟏𝟏𝟐𝟐𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐)

𝑺𝑺𝑨𝑨 = 𝟗𝟗𝟕𝟕.𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 + 𝟏𝟏𝟗𝟗𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 𝑩𝑩 = 𝟏𝟏𝟐𝟐 (𝟏𝟏𝟗𝟗𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐)

𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟒𝟒𝟕𝟕.𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 𝑩𝑩 = 𝟗𝟗𝟓𝟓 𝐜𝐜𝐩𝐩𝟐𝟐

The surface area of the prism is 𝟐𝟐𝟒𝟒𝟕𝟕.𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐.

4. A cube has a volume of 𝟔𝟔𝟒𝟒 𝐩𝐩𝟑𝟑. What is the cube’s surface area?

A cube’s length, width, and height must be equal. 𝟔𝟔𝟒𝟒 = 𝟒𝟒 ∙ 𝟒𝟒 ∙ 𝟒𝟒 = 𝟒𝟒𝟑𝟑, so the length, width, and height of the cube are all 𝟒𝟒 𝐩𝐩.

𝑺𝑺𝑨𝑨 = 𝟔𝟔𝒔𝒔𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟔𝟔(𝟒𝟒 𝐩𝐩)𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟔𝟔(𝟏𝟏𝟔𝟔 𝐩𝐩𝟐𝟐)

𝑺𝑺𝑨𝑨 = 𝟗𝟗𝟔𝟔 𝐩𝐩𝟐𝟐

5. The height of a right rectangular prism is 𝟒𝟒𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩. The length and width of the prism’s base are 𝟐𝟐 𝐟𝐟𝐩𝐩. and 𝟏𝟏𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩. Use the formula 𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨 + 𝟐𝟐𝑩𝑩 to find the surface area of the right rectangular prism.


𝑳𝑳𝑨𝑨 = 𝑷𝑷 ∙ 𝟐𝟐 𝑩𝑩 = 𝒍𝒍𝒍𝒍

𝑳𝑳𝑨𝑨 = �𝟐𝟐 𝐟𝐟𝐩𝐩. +𝟐𝟐 𝐟𝐟𝐩𝐩. +𝟏𝟏𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩. +𝟏𝟏𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩. � ∙ 𝟒𝟒 𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩. 𝑩𝑩 = 𝟐𝟐 𝐟𝐟𝐩𝐩.∙ 𝟏𝟏 𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩.

𝑳𝑳𝑨𝑨 = (𝟐𝟐 𝐟𝐟𝐩𝐩. + 𝟐𝟐 𝐟𝐟𝐩𝐩. +𝟑𝟑 𝐟𝐟𝐩𝐩. ) ∙ 𝟒𝟒 𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩. 𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟐𝟐𝒃𝒃 𝑩𝑩 = 𝟑𝟑 𝐟𝐟𝐩𝐩𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟕𝟕 𝐟𝐟𝐩𝐩.∙ 𝟒𝟒 𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩. 𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟏𝟏𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩𝟐𝟐 + 𝟐𝟐(𝟑𝟑 𝐟𝐟𝐩𝐩𝟐𝟐)

𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟒𝟒 𝐟𝐟𝐩𝐩𝟐𝟐 + 𝟑𝟑𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩𝟐𝟐 𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟏𝟏𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩𝟐𝟐 + 𝟔𝟔 𝐟𝐟𝐩𝐩𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟑𝟑𝟏𝟏𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩𝟐𝟐 𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟕𝟕𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩𝟐𝟐

The surface area of the right rectangular prism is 𝟑𝟑𝟕𝟕𝟏𝟏𝟐𝟐 𝐟𝐟𝐩𝐩𝟐𝟐.

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6. The surface area of a right rectangular prism is 𝟔𝟔𝟒𝟒𝟐𝟐𝟑𝟑 𝐢𝐢𝐢𝐢𝟐𝟐. The dimensions of its base are 𝟑𝟑 𝐢𝐢𝐢𝐢 and 𝟕𝟕 𝐢𝐢𝐢𝐢 Use the formula 𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟐𝟐𝑩𝑩 and 𝑳𝑳𝑨𝑨 = 𝑷𝑷𝟐𝟐 to find the unknown height 𝟐𝟐 of the prism.

𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟐𝟐B

𝑺𝑺𝑨𝑨 = 𝑷𝑷 ∙ 𝟐𝟐 + 𝟐𝟐𝑩𝑩

𝟔𝟔𝟒𝟒𝟐𝟐𝟑𝟑 𝐢𝐢𝐢𝐢𝟐𝟐 = 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢.∙ (𝟐𝟐) + 𝟐𝟐(𝟐𝟐𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐)

𝟔𝟔𝟒𝟒𝟐𝟐𝟑𝟑 𝐢𝐢𝐢𝐢𝟐𝟐 = 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢.∙ (𝟐𝟐) + 𝟒𝟒𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐

𝟔𝟔𝟒𝟒𝟐𝟐𝟑𝟑 𝐢𝐢𝐢𝐢𝟐𝟐 − 𝟒𝟒𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 = 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢.∙ (𝟐𝟐) + 𝟒𝟒𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐 − 𝟒𝟒𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐

𝟐𝟐𝟔𝟔𝟐𝟐𝟑𝟑 𝐢𝐢𝐢𝐢𝟐𝟐 = 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢.∙ (𝟐𝟐) + 𝟐𝟐 𝐢𝐢𝐢𝐢𝟐𝟐

𝟐𝟐𝟔𝟔𝟐𝟐𝟑𝟑 𝐢𝐢𝐢𝐢𝟐𝟐 ∙ 𝟏𝟏𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢. = 𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢 ∙ 𝟏𝟏

𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢. ∙ (𝟐𝟐)

𝟒𝟒𝟐𝟐𝟑𝟑

𝐢𝐢𝐢𝐢𝟐𝟐 ∙𝟏𝟏

𝟐𝟐𝟐𝟐 𝐢𝐢𝐢𝐢= 𝟏𝟏 ∙ 𝟐𝟐

𝟒𝟒𝟑𝟑

𝐢𝐢𝐢𝐢. = 𝟐𝟐

𝟐𝟐 = 𝟒𝟒𝟑𝟑 𝐢𝐢𝐢𝐢. or 𝟏𝟏𝟏𝟏𝟑𝟑 𝐢𝐢𝐢𝐢.

The height of the prism is 𝟏𝟏𝟏𝟏𝟑𝟑 𝐢𝐢𝐢𝐢.

7. A given right triangular prism has an equilateral triangular base. The height of that equilateral triangle is

approximately 𝟕𝟕.𝟏𝟏 𝐜𝐜𝐩𝐩. The distance between the bases is 𝟗𝟗 𝐜𝐜𝐩𝐩. The surface area of the prism is 𝟑𝟑𝟏𝟏𝟗𝟗𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐. Find the approximate lengths of the sides of the base.

𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟐𝟐𝑩𝑩 Let 𝒙𝒙 represent the number of centimeters in each side of the equilateral triangle.

𝑳𝑳𝑨𝑨 = 𝑷𝑷 ∙ 𝟐𝟐 𝑩𝑩 = 𝟏𝟏𝟐𝟐 𝒍𝒍𝒍𝒍 𝟑𝟑𝟏𝟏𝟗𝟗𝟏𝟏𝟐𝟐 𝒄𝒄𝒎𝒎𝟐𝟐 = 𝑳𝑳𝑨𝑨 + 𝟐𝟐𝑩𝑩

𝑳𝑳𝑨𝑨 = 𝟑𝟑(𝒙𝒙 𝐜𝐜𝐩𝐩) ∙ 𝟗𝟗 𝐜𝐜𝐩𝐩 𝑩𝑩 = 𝟏𝟏𝟐𝟐 ∙ (𝒙𝒙 𝐜𝐜𝐩𝐩) ∙ 𝟕𝟕.𝟏𝟏 𝐜𝐜𝐩𝐩 𝟑𝟑𝟏𝟏𝟗𝟗𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 = 𝟐𝟐𝟕𝟕𝒙𝒙 𝐜𝐜𝐩𝐩𝟐𝟐 + 𝟐𝟐(𝟑𝟑.𝟓𝟓𝟓𝟓𝒙𝒙 𝐜𝐜𝐩𝐩𝟐𝟐)

𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟕𝟕𝒙𝒙 𝐜𝐜𝐩𝐩𝟐𝟐 𝑩𝑩 = 𝟑𝟑.𝟓𝟓𝟓𝟓𝒙𝒙 𝐜𝐜𝐩𝐩𝟐𝟐 𝟑𝟑𝟏𝟏𝟗𝟗𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 = 𝟐𝟐𝟕𝟕𝒙𝒙 𝐜𝐜𝐩𝐩𝟐𝟐 + 𝟕𝟕.𝟏𝟏𝒙𝒙 𝐜𝐜𝐩𝐩𝟐𝟐

𝟑𝟑𝟏𝟏𝟗𝟗𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 = 𝟑𝟑𝟒𝟒.𝟏𝟏𝒙𝒙 𝐜𝐜𝐩𝐩𝟐𝟐

𝟑𝟑𝟏𝟏𝟗𝟗𝟏𝟏𝟐𝟐 𝐜𝐜𝐩𝐩𝟐𝟐 = 𝟑𝟑𝟒𝟒 𝟏𝟏𝟏𝟏𝟐𝟐𝒙𝒙 𝐜𝐜𝐩𝐩𝟐𝟐

𝟔𝟔𝟑𝟑𝟗𝟗𝟐𝟐

𝐜𝐜𝐩𝐩𝟐𝟐 =𝟑𝟑𝟒𝟒𝟏𝟏𝟏𝟏𝟐𝟐

𝒙𝒙 𝐜𝐜𝐩𝐩𝟐𝟐

𝟔𝟔𝟑𝟑𝟗𝟗𝟐𝟐

𝐜𝐜𝐩𝐩𝟐𝟐 ∙𝟏𝟏𝟐𝟐

𝟑𝟑𝟒𝟒𝟏𝟏 𝐜𝐜𝐩𝐩=𝟑𝟑𝟒𝟒𝟏𝟏𝟏𝟏𝟐𝟐

𝒙𝒙 𝐜𝐜𝐩𝐩𝟐𝟐 ∙𝟏𝟏𝟐𝟐

𝟑𝟑𝟒𝟒𝟏𝟏 𝐜𝐜𝐩𝐩

𝟑𝟑𝟏𝟏𝟗𝟗𝟓𝟓𝟑𝟑𝟒𝟒𝟏𝟏

𝐜𝐜𝐩𝐩 = 𝒙𝒙

𝒙𝒙 =𝟑𝟑𝟏𝟏𝟗𝟗𝟓𝟓𝟑𝟑𝟒𝟒𝟏𝟏

𝐜𝐜𝐩𝐩

𝒙𝒙 ≈ 𝟗𝟗.𝟒𝟒 𝐜𝐜𝐩𝐩

The lengths of the sides of the equilateral triangles are approximately 𝟗𝟗.𝟒𝟒 𝐜𝐜𝐩𝐩 each.

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Student Outcomes

Students find the surface area of three-dimensional objects whose surface area is composed of triangles and quadrilaterals, specifically focusing on pyramids. They use polyhedron nets to understand that surface area is simply the sum of the area of the lateral faces and the area of the base(s).

Lesson Notes Before class, teachers need to make copies of the composite figure for the Opening Exercise on cardstock. To save class time, they could also cut these nets out for students.

Classwork


Make copies of the composite figure on cardstock and have students cut and fold the net to form the three-dimensional object.

Opening Exercise

What is the area of the composite figure in the diagram? Is the diagram a net for a three-dimensional image? If so, sketch the image. If not, explain why.

There are four unit squares in each square of the figure. There are 𝟏𝟏𝟏𝟏 total squares that make up the figure, so the total area of the composite figure is

𝑨𝑨 = 𝟏𝟏𝟏𝟏 ∙ 𝟒𝟒 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐬𝐬𝟐𝟐 = 𝟕𝟕𝟐𝟐 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐬𝐬𝟐𝟐.

The composite figure does represent the net of a three-dimensional figure. The figure is shown below.

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Pyramids are formally defined and explored in more depth in Module 6. Here, we simply introduce finding the surface area of a pyramid and ask questions designed to elicit the formulas from students. For example, ask how many lateral faces there are on the pyramid; then, ask for the triangle area formula. Continue leading students toward stating the formula for total surface area on their own.

Example 1

The pyramid in the picture has a square base, and its lateral faces are triangles that are exact copies of one another. Find the surface area of the pyramid.

The surface area of the pyramid consists of one square base and four lateral triangular faces.

𝑳𝑳𝑨𝑨 = 𝟒𝟒�𝟏𝟏𝟐𝟐𝒃𝒃𝒃𝒃� 𝑩𝑩 = 𝒔𝒔𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟒𝟒 ∙ 𝟏𝟏𝟐𝟐 (𝟔𝟔 𝐜𝐜𝐜𝐜 ∙ 𝟕𝟕 𝐜𝐜𝐜𝐜) 𝑩𝑩 = (𝟔𝟔 𝐜𝐜𝐜𝐜)𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟐𝟐(𝟔𝟔 𝐜𝐜𝐜𝐜 ∙ 𝟕𝟕 𝐜𝐜𝐜𝐜) 𝑩𝑩 = 𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟐𝟐(𝟒𝟒𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐) The pyramid’s base area is 𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐.

𝑳𝑳𝑨𝑨 = 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟐𝟐

The pyramid’s lateral area is 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟐𝟐. 𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝑩𝑩

𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

The surface area of the pyramid is 𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐.

Example 2 (4 minutes): Using Cubes

Consider providing 13 interlocking cubes to small groups of students so they may construct a model of the diagram shown. Remind students to count faces systematically. For example, first consider only the bottom 9 cubes. This structure has a surface area of 30 (9 at the top, 9 at the bottom, and 3 on each of the four sides). Now consider the four cubes added at the top. Since we have already counted the tops of these cubes, we just need to add the four sides of

each. 30 + 16 = 46 total square faces, each with side length 14 inch.

Example 2: Using Cubes

There are 𝟏𝟏𝟑𝟑 cubes glued together forming the solid in the diagram. The edges of each cube are 𝟏𝟏𝟒𝟒 inch in length. Find the surface area of the solid.

The surface area of the solid consists of 𝟒𝟒𝟔𝟔 square faces, all having side lengths of 𝟏𝟏𝟒𝟒 inch. The area of a square having

sides of length 𝟏𝟏𝟒𝟒 inch is 𝟏𝟏𝟏𝟏𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐 .

𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟔𝟔 ∙ 𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮𝐬𝐬𝐬𝐬𝐬𝐬

𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟔𝟔 ∙𝟏𝟏𝟏𝟏𝟔𝟔

𝐮𝐮𝐮𝐮𝟐𝟐

𝑺𝑺𝑨𝑨 =𝟒𝟒𝟔𝟔𝟏𝟏𝟔𝟔


𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟏𝟏𝟒𝟒𝟏𝟏𝟔𝟔


𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟕𝟕𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 The surface are of the solid is 𝟐𝟐𝟕𝟕𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐.

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Compare and contrast the methods for finding surface area for pyramids and prisms. How are the methods similar?

When finding the surface area of a pyramid and a prism, we find the area of each face and then add these areas together.

How are they different? Calculating the surface area of pyramids and prisms is different because of the shape of the faces. A

prism has two bases and lateral faces that are rectangles. A pyramid has only one base and lateral faces that are triangles.


Example 3

Find the total surface area of the wooden jewelry box. The sides and bottom of the box are all 𝟏𝟏𝟒𝟒 inch thick. What are the faces that make up this box?

The box has a rectangular bottom, rectangular lateral faces, and a rectangular top that has a smaller rectangle removed from it. There are also rectangular faces that make up the inner lateral faces and the inner bottom of the box.

How does this box compare to other objects that you have found the surface area of?

The box is a rectangular prism with a smaller rectangular prism removed from its inside. The total surface area is equal to the surface area of the larger right rectangular prism plus the lateral area of the smaller right rectangular prism.

Large Prism: The surface area of the large right rectangular prism makes up the outside faces of the box, the rim of the box, and the inside bottom face of the box. 𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟐𝟐𝑩𝑩 𝑳𝑳𝑨𝑨 = 𝑷𝑷 ∙ 𝒃𝒃 𝑩𝑩 = 𝒍𝒍𝒍𝒍 𝑳𝑳𝑨𝑨 = 𝟑𝟑𝟐𝟐 𝐮𝐮𝐮𝐮.∙ 𝟒𝟒 𝐮𝐮𝐮𝐮. 𝑩𝑩 = 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮.∙ 𝟔𝟔 𝐮𝐮𝐮𝐮. 𝑳𝑳𝑨𝑨 = 𝟏𝟏𝟐𝟐𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 𝑩𝑩 = 𝟔𝟔𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 The lateral area is 𝟏𝟏𝟐𝟐𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐. The base area is 𝟔𝟔𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐. 𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟐𝟐𝑩𝑩 𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟐𝟐𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟐𝟐(𝟔𝟔𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐) 𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟐𝟐𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏𝟐𝟐𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟒𝟒𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 The surface area of the larger prism is 𝟐𝟐𝟒𝟒𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐.

Small Prism: The smaller prism is 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. smaller in

length and width and 𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮. smaller in height due to the thickness of the sides of the box.

𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟏𝟏𝑩𝑩 𝑳𝑳𝑨𝑨 = 𝑷𝑷 ∙ 𝒃𝒃

𝑳𝑳𝑨𝑨 = 𝟐𝟐�𝟗𝟗𝟏𝟏𝟐𝟐

𝐮𝐮𝐮𝐮. +𝟓𝟓𝟏𝟏𝟐𝟐

𝐮𝐮𝐮𝐮. � ∙ 𝟑𝟑𝟑𝟑𝟒𝟒

𝐮𝐮𝐮𝐮.

𝑳𝑳𝑨𝑨 = 𝟐𝟐(𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮. +𝟏𝟏 𝐮𝐮𝐮𝐮. ) ∙ 𝟑𝟑𝟑𝟑𝟒𝟒

𝐮𝐮𝐮𝐮.

𝑳𝑳𝑨𝑨 = 𝟐𝟐(𝟏𝟏𝟓𝟓 𝐮𝐮𝐮𝐮. ) ∙ 𝟑𝟑𝟑𝟑𝟒𝟒

𝐮𝐮𝐮𝐮.

𝑳𝑳𝑨𝑨 = 𝟑𝟑𝟏𝟏 𝐮𝐮𝐮𝐮.∙ 𝟑𝟑𝟑𝟑𝟒𝟒

𝐮𝐮𝐮𝐮.

𝑳𝑳𝑨𝑨 = 𝟗𝟗𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 +𝟗𝟗𝟏𝟏𝟒𝟒


𝑳𝑳𝑨𝑨 = 𝟗𝟗𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟏𝟏𝟐𝟐


𝑳𝑳𝑨𝑨 = 𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐


The lateral area is 𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐.

Surface Area of the Box

𝑺𝑺𝑨𝑨𝐛𝐛𝐛𝐛𝐛𝐛 = 𝑺𝑺𝑨𝑨 + 𝑳𝑳𝑨𝑨

𝑺𝑺𝑨𝑨𝒃𝒃𝒃𝒃𝒃𝒃 = 𝟐𝟐𝟒𝟒𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐 𝒊𝒊𝒏𝒏𝟐𝟐

𝑺𝑺𝑨𝑨𝒃𝒃𝒃𝒃𝒃𝒃 = 𝟑𝟑𝟔𝟔𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐

The total surface area of the box is 𝟑𝟑𝟔𝟔𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐.

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Discussion (5 minutes): Strategies and Observations from Example 3

Call on students to provide their answers to each of the following questions. To encourage a discussion about strategy, patterns, arguments, or observations, call on more than one student for each of these questions.

What ideas did you have to solve this problem? Explain.

Answers will vary.

Did you make any mistakes in your solution? Explain.

Answers will vary; examples include the following:

I subtracted 12 inch from the depth of the box instead of 14 inch;

I subtracted only 14 inch from the length and width because I didn’t account for both sides.

Describe how you found the surface area of the box and what that surface area is. Answers will vary.

Closing (2 minutes)

What are some strategies for finding the surface area of solids? Answers will vary but might include creating nets, adding the areas of polygonal faces, and counting

square faces and adding their areas.


Scaffolding: To help students visualize the various faces involved on this object, consider constructing a similar object by placing a smaller shoe box inside a slightly larger shoe box. This will also help students visualize the inner surfaces of the box as the lateral faces of the smaller prism that is removed from the larger prism.

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Name Date


Exit Ticket 1. The right hexagonal pyramid has a hexagon base with equal-length sides. The

lateral faces of the pyramid are all triangles (that are exact copies of one another) with heights of 15 ft. Find the surface area of the pyramid.

2. Six cubes are glued together to form the solid shown in the

diagram. If the edges of each cube measure 1 12 inches in length,

what is the surface area of the solid?

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1. The right hexagonal pyramid has a hexagon base with equal-length sides. The lateral faces of the pyramid are all triangles (that are exact copies of one another) with heights of 𝟏𝟏𝟓𝟓 𝐟𝐟𝐮𝐮. Find the surface area of the pyramid.

𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟏𝟏𝑩𝑩

𝑳𝑳𝑨𝑨 = 𝟔𝟔 ∙ 𝟏𝟏𝟐𝟐 (𝒃𝒃𝒃𝒃) 𝑩𝑩 = 𝑨𝑨rectangle + 𝟐𝟐𝑨𝑨triangle

𝑳𝑳𝑨𝑨 = 𝟔𝟔 ∙ 𝟏𝟏𝟐𝟐 (𝟓𝟓 𝐟𝐟𝐮𝐮.∙ 𝟏𝟏𝟓𝟓 𝐟𝐟𝐮𝐮. ) 𝑩𝑩 = (𝟏𝟏 𝒇𝒇𝒇𝒇.∙ 𝟓𝟓 𝒇𝒇𝒇𝒇. ) + 𝟐𝟐 ∙ 𝟏𝟏𝟐𝟐 (𝟏𝟏 𝒇𝒇𝒇𝒇.∙ 𝟑𝟑 𝒇𝒇𝒇𝒇. )

𝑳𝑳𝑨𝑨 = 𝟑𝟑 ∙ 𝟕𝟕𝟓𝟓 𝐟𝐟𝐮𝐮𝟐𝟐 𝑩𝑩 = 𝟒𝟒𝟏𝟏 𝐟𝐟𝐮𝐮𝟐𝟐 + (𝟏𝟏 𝐟𝐟𝐮𝐮.∙ 𝟑𝟑 𝐟𝐟𝐮𝐮. )

𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟐𝟐𝟓𝟓 𝐟𝐟𝐮𝐮𝟐𝟐 𝑩𝑩 = 𝟒𝟒𝟏𝟏 𝐟𝐟𝐮𝐮𝟐𝟐 + 𝟐𝟐𝟒𝟒 𝐟𝐟𝐮𝐮𝟐𝟐

𝑩𝑩 = 𝟔𝟔𝟒𝟒 𝐟𝐟𝐮𝐮𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟏𝟏𝑩𝑩

𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟐𝟐𝟓𝟓 𝐟𝐟𝐮𝐮𝟐𝟐 + 𝟔𝟔𝟒𝟒 𝐟𝐟𝐮𝐮𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟏𝟏𝟗𝟗 𝐟𝐟𝐮𝐮𝟐𝟐

The surface area of the pyramid is 𝟐𝟐𝟏𝟏𝟗𝟗 𝐟𝐟𝐮𝐮𝟐𝟐.

2. Six cubes are glued together to form the solid shown in the diagram. If the edges of each cube measure 𝟏𝟏𝟏𝟏𝟐𝟐 inches in length, what is the surface area of the solid?

There are 𝟐𝟐𝟔𝟔 square cube faces showing on the surface area of the solid (𝟓𝟓 each from the top and bottom view, 𝟒𝟒 each from the front and back view, 𝟑𝟑 each from the left and right side views, and 𝟐𝟐 from the “inside” of the front).

𝑨𝑨 = 𝒔𝒔𝟐𝟐 𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟔𝟔 ∙ �𝟐𝟐𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐�

𝑨𝑨 = �𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. �𝟐𝟐 𝑺𝑺𝑨𝑨 = 𝟓𝟓𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟐𝟐𝟔𝟔

𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐

𝑨𝑨 = �𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. � �𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. � 𝑺𝑺𝑨𝑨 = 𝟓𝟓𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐

𝑨𝑨 = 𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. �𝟏𝟏 𝐮𝐮𝐮𝐮. +𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. � 𝑺𝑺𝑨𝑨 = 𝟓𝟓𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐

𝑨𝑨 = �𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮.∙ 𝟏𝟏 𝐮𝐮𝐮𝐮. � + �𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮.∙ 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. �

𝑨𝑨 = 𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐

𝑨𝑨 = 𝟏𝟏𝟐𝟐𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐

𝑨𝑨 = 𝟏𝟏𝟓𝟓𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐 = 𝟐𝟐𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐 The surface area of the solid is 𝟓𝟓𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐.

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7•3 Lesson 22



1. For each of the following nets, draw (or describe) the solid represented by the net and find its surface area.

a. The equilateral triangles are exact copies.

The net represents a triangular pyramid where the three lateral faces are identical to each other and the triangular base. 𝑺𝑺𝑨𝑨 = 𝟒𝟒𝑩𝑩 since the faces are all the same size and shape.

𝑩𝑩 = 𝟏𝟏𝟐𝟐𝒃𝒃𝒃𝒃 𝑺𝑺𝑨𝑨 = 𝟒𝟒𝑩𝑩

𝑩𝑩 = 𝟏𝟏𝟐𝟐 ∙ 𝟗𝟗 𝐜𝐜𝐜𝐜 ∙ 𝟕𝟕𝟒𝟒𝟓𝟓 𝐜𝐜𝐜𝐜 𝑺𝑺𝑨𝑨 = 𝟒𝟒�𝟑𝟑𝟓𝟓 𝟏𝟏

𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐�

𝑩𝑩 = 𝟗𝟗𝟐𝟐 𝐜𝐜𝐜𝐜 ∙ 𝟕𝟕𝟒𝟒𝟓𝟓 𝐜𝐜𝐜𝐜 𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟒𝟒𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟒𝟒

𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

𝑩𝑩 = 𝟔𝟔𝟑𝟑𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟑𝟑𝟔𝟔

𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟒𝟒𝟏𝟏𝟐𝟐𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐

𝑩𝑩 = 𝟑𝟑𝟏𝟏𝟓𝟓𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟑𝟑𝟔𝟔

𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

𝑩𝑩 = 𝟑𝟑𝟓𝟓𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 The surface area of the triangular pyramid is 𝟏𝟏𝟒𝟒𝟏𝟏𝟐𝟐𝟓𝟓 𝐜𝐜𝐜𝐜𝟐𝟐.

𝑩𝑩 = 𝟑𝟑𝟓𝟓𝟏𝟏𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜𝟐𝟐

b. The net represents a square pyramid that has four identical lateral faces that are triangles. The base is a square.

𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝑩𝑩

𝑳𝑳𝑨𝑨 = 𝟒𝟒 ∙ 𝟏𝟏𝟐𝟐 (𝒃𝒃𝒃𝒃) 𝑩𝑩 = 𝒔𝒔𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟒𝟒 ∙ 𝟏𝟏𝟐𝟐 �𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮.∙ 𝟏𝟏𝟒𝟒 𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮. � 𝑩𝑩 = (𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮.)𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟐𝟐�𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮.∙ 𝟏𝟏𝟒𝟒 𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮.� 𝑩𝑩 = 𝟏𝟏𝟒𝟒𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟐𝟐(𝟏𝟏𝟔𝟔𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟗𝟗 𝐮𝐮𝐮𝐮𝟐𝟐)

𝑳𝑳𝑨𝑨 = 𝟑𝟑𝟑𝟑𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟑𝟑𝟓𝟓𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝑩𝑩

𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟓𝟓𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏𝟒𝟒𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟗𝟗𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐

The surface area of the square pyramid is 𝟒𝟒𝟗𝟗𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐.

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7•3 Lesson 22


2. Find the surface area of the following prism.


𝑳𝑳𝑨𝑨 = 𝑷𝑷 ∙ 𝒃𝒃

𝑳𝑳𝑨𝑨 = �𝟒𝟒 𝐜𝐜𝐜𝐜+ 𝟔𝟔𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜 + 𝟒𝟒𝟏𝟏𝟓𝟓 𝐜𝐜𝐜𝐜 + 𝟓𝟓𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜� ∙ 𝟗𝟗 𝐜𝐜𝐜𝐜

𝑳𝑳𝑨𝑨 = �𝟏𝟏𝟗𝟗 𝐜𝐜𝐜𝐜+ 𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜 + 𝟏𝟏

𝟓𝟓 𝐜𝐜𝐜𝐜 + 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜� ∙ 𝟗𝟗 𝐜𝐜𝐜𝐜

𝑳𝑳𝑨𝑨 = �𝟏𝟏𝟗𝟗 𝐜𝐜𝐜𝐜+ 𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜 + 𝟒𝟒

𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜 + 𝟓𝟓𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜� ∙ 𝟗𝟗 𝐜𝐜𝐜𝐜

𝑳𝑳𝑨𝑨 = �𝟏𝟏𝟗𝟗 𝐜𝐜𝐜𝐜+ 𝟏𝟏𝟗𝟗𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜� ∙ 𝟗𝟗 𝐜𝐜𝐜𝐜

𝑳𝑳𝑨𝑨 = 𝟏𝟏𝟕𝟕𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟏𝟏𝟕𝟕𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟏𝟏𝟕𝟕𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

𝑳𝑳𝑨𝑨 = 𝟏𝟏𝟕𝟕𝟗𝟗𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

𝑩𝑩 = 𝑨𝑨𝐬𝐬𝐬𝐬𝐜𝐜𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 + 𝑨𝑨𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟐𝟐𝑩𝑩

𝑩𝑩 = �𝟓𝟓𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜 ∙ 𝟒𝟒 𝐜𝐜𝐜𝐜�+ 𝟏𝟏𝟐𝟐�𝟒𝟒 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜� 𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟕𝟕𝟗𝟗𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟐𝟐�𝟐𝟐𝟑𝟑𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐�

𝑩𝑩 = (𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐) + �𝟐𝟐 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜� 𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟕𝟕𝟗𝟗𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟒𝟒𝟕𝟕 𝐜𝐜𝐜𝐜𝟐𝟐

𝑩𝑩 = 𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟐𝟐𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐 𝐒𝐒𝐒𝐒 = 𝟐𝟐𝟐𝟐𝟔𝟔𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

𝑩𝑩 = 𝟐𝟐𝟑𝟑𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐

The surface area of the prism is 𝟐𝟐𝟐𝟐𝟔𝟔𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐.

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7•3 Lesson 22


3. The net below is for a specific object. The measurements shown are in meters. Sketch (or describe) the object, and then find its surface area.


𝑳𝑳𝑨𝑨 = 𝑷𝑷 ∙ 𝒃𝒃 𝑩𝑩 = �𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜� + �𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏 𝐜𝐜𝐜𝐜�+ �𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜� 𝑺𝑺𝑨𝑨 = 𝑳𝑳𝑨𝑨+ 𝟐𝟐𝑩𝑩

𝑳𝑳𝑨𝑨 = 𝟔𝟔 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜 𝑩𝑩 = �𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟐𝟐�+ �𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐� + �𝟑𝟑𝟒𝟒 𝐜𝐜𝐜𝐜𝟐𝟐� 𝑺𝑺𝑨𝑨 = 𝟑𝟑 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟐𝟐�𝟏𝟏𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐�

𝑳𝑳𝑨𝑨 = 𝟑𝟑 𝐜𝐜𝐜𝐜𝟐𝟐 𝑩𝑩 = �𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟐𝟐�+ �𝟐𝟐𝟒𝟒 𝐜𝐜𝐜𝐜𝟐𝟐� + �𝟑𝟑𝟒𝟒 𝐜𝐜𝐜𝐜𝟐𝟐� 𝑺𝑺𝑨𝑨 = 𝟑𝟑 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟑𝟑 𝐜𝐜𝐜𝐜𝟐𝟐

𝑩𝑩 = 𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜𝟐𝟐 𝑺𝑺𝑨𝑨 = 𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐

𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐

The surface area of the object is 𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐.

4. In the diagram, there are 𝟏𝟏𝟒𝟒 cubes glued together to form a solid. Each cube has a volume of 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟑𝟑. Find the surface area of the solid.

The volume of a cube is 𝒔𝒔𝟑𝟑, and 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟑𝟑 is the same as �𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. �𝟑𝟑, so the

cubes have edges that are 𝟏𝟏𝟐𝟐𝐮𝐮𝐮𝐮. long. The cube faces have area 𝒔𝒔𝟐𝟐, or

�𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. �𝟐𝟐, or

𝟏𝟏𝟒𝟒

𝐮𝐮𝐮𝐮𝟐𝟐. There are 𝟒𝟒𝟐𝟐 cube faces that make up the surface

of the solid.

𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐 ∙ 𝟒𝟒𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐

The surface area of the solid is 𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐.


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7•3 Lesson 22


c. a. b.

5. The nets below represent three solids. Sketch (or describe) each solid, and find its surface area.



𝑳𝑳𝑨𝑨 = 𝟏𝟏𝟐𝟐 ∙ 𝟑𝟑

𝑳𝑳𝑨𝑨 = 𝟑𝟑𝟔𝟔 𝒄𝒄𝒄𝒄𝟐𝟐

𝑩𝑩 = 𝒔𝒔𝟐𝟐

𝑩𝑩 = (𝟑𝟑 𝐜𝐜𝐜𝐜)𝟐𝟐

𝑩𝑩 = 𝟗𝟗 𝐜𝐜𝐜𝐜𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟐𝟐(𝟗𝟗 𝐜𝐜𝐜𝐜𝟐𝟐) 𝑺𝑺𝑨𝑨 = 𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 𝑺𝑺𝑨𝑨 = 𝟓𝟓𝟒𝟒 𝐜𝐜𝐜𝐜𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟑𝟑𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮𝐬𝐬𝐬𝐬𝐬𝐬 + 𝟑𝟑𝑨𝑨𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 + 𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬

𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮𝐬𝐬𝐬𝐬𝐬𝐬 = 𝒔𝒔𝟐𝟐

𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮𝐬𝐬𝐬𝐬𝐬𝐬 = (𝟑𝟑 𝐜𝐜𝐜𝐜)𝟐𝟐

𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮𝐬𝐬𝐬𝐬𝐬𝐬 = 𝟗𝟗 𝐜𝐜𝐜𝐜𝟐𝟐

𝑨𝑨𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 =𝟏𝟏𝟐𝟐𝒃𝒃𝒃𝒃

𝑨𝑨𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 =𝟏𝟏𝟐𝟐∙ 𝟑𝟑 𝐜𝐜𝐜𝐜 ∙ 𝟑𝟑 𝐜𝐜𝐜𝐜

𝑨𝑨𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 =𝟗𝟗𝟐𝟐

𝑨𝑨𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐬𝐬 = 𝟒𝟒𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜2

𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 =𝟏𝟏𝟐𝟐𝒃𝒃𝒃𝒃

𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 =𝟏𝟏𝟐𝟐∙ �𝟒𝟒

𝟏𝟏𝟓𝟓

𝐜𝐜𝐜𝐜� ∙ �𝟑𝟑𝟕𝟕𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜�

𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 = 𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜 ∙ 𝟑𝟑𝟕𝟕𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜

𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 =𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜 ∙𝟑𝟑𝟕𝟕𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜

𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 =𝟕𝟕𝟕𝟕𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 = 𝟕𝟕𝟕𝟕𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟑𝟑(𝟗𝟗 𝐜𝐜𝐜𝐜𝟐𝟐) + 𝟑𝟑�𝟒𝟒𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜𝟐𝟐�+ 𝟕𝟕𝟕𝟕𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟕𝟕 𝐜𝐜𝐜𝐜𝟐𝟐 + �𝟏𝟏𝟐𝟐 +𝟑𝟑𝟐𝟐� 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟕𝟕

𝟕𝟕𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟕𝟕 𝐜𝐜𝐜𝐜𝟐𝟐 +𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜𝟐𝟐 +𝟕𝟕𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟕𝟕 𝐜𝐜𝐜𝐜𝟐𝟐 +𝟓𝟓𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜𝟐𝟐 +𝟕𝟕𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟕𝟕 𝐜𝐜𝐜𝐜𝟐𝟐 +𝟏𝟏𝟐𝟐𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟑𝟑𝑨𝑨𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬 + 𝑨𝑨𝐬𝐬𝐬𝐬𝐮𝐮 𝐮𝐮𝐬𝐬𝐮𝐮𝐬𝐬𝐮𝐮𝐚𝐚𝐚𝐚𝐬𝐬

𝑺𝑺𝑨𝑨 = 𝟑𝟑�𝟒𝟒𝟏𝟏𝟐𝟐� 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟕𝟕

𝟕𝟕𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟏𝟏𝟐𝟐 𝐜𝐜𝐜𝐜𝟐𝟐 +𝟑𝟑𝟐𝟐

𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟕𝟕 𝐜𝐜𝐜𝐜𝟐𝟐 +𝟕𝟕𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 +𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜𝟐𝟐 +𝟕𝟕𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 +𝟐𝟐𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟏𝟏𝟐𝟐𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜2

𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟕𝟕 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 +𝟐𝟐𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟏𝟏𝟐𝟐𝟕𝟕𝟏𝟏𝟏𝟏𝟏𝟏


Sketch B

Sketch C

Sketch A

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317




7•3 Lesson 22


d. How are figures (b) and (c) related to figure (a)?

If the equilateral triangular faces of figures (b) and (c) were matched together, they would form the cube in part (a).

6. Find the surface area of the solid shown in the diagram. The solid is a right triangular prism (with right triangular bases) with a smaller right triangular prism removed from it.



𝑳𝑳𝑨𝑨 = �𝟒𝟒 𝐮𝐮𝐮𝐮. +𝟒𝟒 𝐮𝐮𝐮𝐮. +𝟓𝟓𝟏𝟏𝟑𝟑𝟐𝟐𝟏𝟏

𝐮𝐮𝐮𝐮. � ∙ 𝟐𝟐 𝐮𝐮𝐮𝐮.

𝑳𝑳𝑨𝑨 = �𝟏𝟏𝟑𝟑𝟏𝟏𝟑𝟑𝟐𝟐𝟏𝟏

𝐮𝐮𝐮𝐮. � ∙ 𝟐𝟐 𝐮𝐮𝐮𝐮.

𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐 +𝟏𝟏𝟑𝟑𝟏𝟏𝟏𝟏


𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 +𝟑𝟑𝟏𝟏𝟏𝟏


𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟕𝟕𝟑𝟑𝟏𝟏𝟏𝟏


The 𝟏𝟏𝟒𝟒

𝐮𝐮𝐮𝐮. by 𝟒𝟒𝟏𝟏𝟗𝟗𝟐𝟐𝟏𝟏 𝐮𝐮𝐮𝐮. rectangle has to be taken away from the lateral area:

𝑨𝑨 = 𝒍𝒍𝒍𝒍 𝐋𝐋𝐒𝐒 = 𝟐𝟐𝟕𝟕 𝟑𝟑𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 − 𝟏𝟏𝟏𝟏𝟗𝟗𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐

𝑨𝑨 = 𝟒𝟒𝟏𝟏𝟗𝟗𝟐𝟐𝟏𝟏 𝐮𝐮𝐮𝐮 ∙ 𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮 𝐋𝐋𝐒𝐒 = 𝟐𝟐𝟕𝟕𝟐𝟐𝟒𝟒𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 − 𝟏𝟏𝟏𝟏𝟗𝟗𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐

𝑨𝑨 = 𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏𝟗𝟗𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟔𝟔 𝟓𝟓

𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐

𝑨𝑨 = 𝟏𝟏𝟏𝟏𝟗𝟗𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 𝑳𝑳𝑨𝑨 = 𝟐𝟐𝟔𝟔 𝟏𝟏𝟏𝟏𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐

Two bases of the larger and smaller triangular prisms must be added:

𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟔𝟔𝟏𝟏𝟏𝟏𝟔𝟔

𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟐𝟐�𝟑𝟑𝟏𝟏𝟐𝟐

𝐮𝐮𝐮𝐮 ∙𝟏𝟏𝟒𝟒

𝐮𝐮𝐮𝐮�+ 𝟐𝟐�𝟏𝟏𝟐𝟐∙ 𝟒𝟒 𝐮𝐮𝐮𝐮 ∙ 𝟒𝟒 𝐮𝐮𝐮𝐮�

𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟔𝟔𝟏𝟏𝟏𝟏𝟔𝟔

𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟐𝟐 ∙𝟏𝟏𝟒𝟒

𝐮𝐮𝐮𝐮 ∙ 𝟑𝟑𝟏𝟏𝟐𝟐

𝐮𝐮𝐮𝐮 + 𝟏𝟏𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟔𝟔𝟏𝟏𝟏𝟏𝟔𝟔

𝐮𝐮𝐮𝐮𝟐𝟐 +𝟏𝟏𝟐𝟐

𝐮𝐮𝐮𝐮 ∙ 𝟑𝟑𝟏𝟏𝟐𝟐

𝐮𝐮𝐮𝐮 + 𝟏𝟏𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟔𝟔𝟏𝟏𝟏𝟏𝟔𝟔

𝐮𝐮𝐮𝐮𝟐𝟐 + �𝟑𝟑𝟐𝟐

𝐮𝐮𝐮𝐮𝟐𝟐 +𝟏𝟏𝟒𝟒

𝐮𝐮𝐮𝐮𝟐𝟐�+ 𝟏𝟏𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟐𝟐𝟔𝟔𝟏𝟏𝟏𝟏𝟔𝟔

𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 +𝟏𝟏𝟏𝟏𝟔𝟔

𝐮𝐮𝐮𝐮𝟐𝟐 +𝟒𝟒𝟏𝟏𝟔𝟔

𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐

𝑺𝑺𝑨𝑨 = 𝟒𝟒𝟑𝟑𝟏𝟏𝟑𝟑𝟏𝟏𝟔𝟔


The surface area of the solid is 𝟒𝟒𝟑𝟑𝟏𝟏𝟑𝟑𝟏𝟏𝟔𝟔 𝐮𝐮𝐮𝐮𝟐𝟐.

A STORY OF RATIOS


318




7•3 Lesson 22


7. The diagram shows a cubic meter that has had three square holes punched completely through the cube on three perpendicular axes. Find the surface area of the remaining solid.

Exterior surfaces of the cube (𝑺𝑺𝑨𝑨𝟏𝟏):

𝑺𝑺𝑨𝑨𝟏𝟏 = 𝟔𝟔(𝟏𝟏 𝐜𝐜)𝟐𝟐 − 𝟔𝟔 �𝟏𝟏𝟐𝟐

𝐜𝐜�𝟐𝟐

𝑺𝑺𝑨𝑨𝟏𝟏 = 𝟔𝟔(𝟏𝟏 𝐜𝐜𝟐𝟐)− 𝟔𝟔�𝟏𝟏𝟒𝟒

𝐜𝐜𝟐𝟐�

𝑺𝑺𝑨𝑨𝟏𝟏 = 𝟔𝟔 𝐜𝐜𝟐𝟐 −𝟔𝟔𝟒𝟒

𝐜𝐜𝟐𝟐

𝑺𝑺𝑨𝑨𝟏𝟏 = 𝟔𝟔 𝐜𝐜𝟐𝟐 − �𝟏𝟏𝟏𝟏𝟐𝟐

𝐜𝐜𝟐𝟐�

𝑺𝑺𝑨𝑨𝟏𝟏 = 𝟒𝟒𝟏𝟏𝟐𝟐

𝐜𝐜𝟐𝟐

Just inside each square hole are four intermediate surfaces that can be treated as the lateral area of a rectangular

prism. Each has a height of 𝟏𝟏𝟒𝟒 𝐜𝐜 and perimeter of 𝟏𝟏𝟐𝟐 𝐜𝐜 + 𝟏𝟏𝟐𝟐 𝐜𝐜 + 𝟏𝟏

𝟐𝟐 𝐜𝐜 + 𝟏𝟏𝟐𝟐 𝐜𝐜 or 𝟐𝟐 𝐜𝐜.

𝑺𝑺𝑨𝑨𝟐𝟐 = 𝟔𝟔(𝑳𝑳𝑨𝑨)

𝑺𝑺𝑨𝑨𝟐𝟐 = 𝟔𝟔�𝟐𝟐 𝐜𝐜 ∙𝟏𝟏𝟒𝟒

𝐜𝐜�

𝑺𝑺𝑨𝑨𝟐𝟐 = 𝟔𝟔 ∙𝟏𝟏𝟐𝟐

𝐜𝐜2

𝑺𝑺𝑨𝑨𝟐𝟐 = 𝟑𝟑 𝐜𝐜𝟐𝟐

The total surface area of the remaining solid is the sum of these two areas:

𝑺𝑺𝑨𝑨𝑻𝑻 = 𝑺𝑺𝑨𝑨𝟏𝟏 + 𝑺𝑺𝑨𝑨𝟐𝟐.

𝑺𝑺𝑨𝑨𝑻𝑻 = 𝟒𝟒𝟏𝟏𝟐𝟐

𝐜𝐜2 + 𝟑𝟑 𝐜𝐜𝟐𝟐

𝑺𝑺𝑨𝑨𝑻𝑻 = 𝟕𝟕𝟏𝟏𝟐𝟐

𝐜𝐜𝟐𝟐

The surface area of the remaining solid is 𝟕𝟕𝟏𝟏𝟐𝟐 𝐜𝐜𝟐𝟐.

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7•3 Lesson 23

Lesson 23: The Volume of a Right Prism


Student Outcomes

Students use the known formula for the volume of a right rectangular prism (length × width × height). Students understand the volume of a right prism to be the area of the base times the height.

Students compute volumes of right prisms involving fractional values for length.

Lesson Notes Students extend their knowledge of obtaining volumes of right rectangular prisms via dimensional measurements to understand how to calculate the volumes of other right prisms. This concept will later be extended to finding the volumes of liquids in right prism-shaped containers and extended again (in Module 6) to finding the volumes of irregular solids using displacement of liquids in containers. The Problem Set scaffolds in the use of equations to calculate unknown dimensions.

Classwork


The volume of a solid is a quantity given by the number of unit cubes needed to fill the solid. Most solids—rocks, baseballs, people—cannot be filled with unit cubes or assembled from cubes. Yet such solids still have volume. Fortunately, we do not need to assemble solids from unit cubes in order to calculate their volume. One of the first interesting examples of a solid that cannot be assembled from cubes, but whose volume can still be calculated from a formula, is a right triangular prism.

What is the area of the square pictured on the right? Explain.

The area of the square is 𝟑𝟑𝟑𝟑 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟐𝟐 because the region is filled with 𝟑𝟑𝟑𝟑 square regions that are 𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 by 𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮, or 𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟐𝟐.

Draw the diagonal joining the two given points; then, darken the grid lines within the lower triangular region. What is the area of that triangular region? Explain.

The area of the triangular region is 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟐𝟐. There are 𝟏𝟏𝟏𝟏 unit squares from the original square and 𝟑𝟑 triangular

regions that are 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟐𝟐. The 𝟑𝟑 triangles can be paired together to form 𝟑𝟑 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟐𝟐. Altogether the area of the triangular region is (𝟏𝟏𝟏𝟏+ 𝟑𝟑) 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟐𝟐, or 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟐𝟐.

How do the areas of the square and the triangular region compare?

The area of the triangular region is half the area of the square region.

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7•3 Lesson 23


Exploratory Challenge (15 minutes): The Volume of a Right Prism

Exploratory Challenge is a continuation of the Opening Exercise.

Exploratory Challenge: The Volume of a Right Prism

What is the volume of the right prism pictured on the right? Explain.

The volume of the right prism is 𝟑𝟑𝟑𝟑 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟑𝟑 because the prism is filled with 𝟑𝟑𝟑𝟑 cubes that are 𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 long, 𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 wide, and 𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 high, or 𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟑𝟑.

Draw the same diagonal on the square base as done above; then, darken the grid lines on the lower right triangular prism. What is the volume of that right triangular prism? Explain.

The volume of the right triangular prism is 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟑𝟑. There are 𝟏𝟏𝟏𝟏 cubes from the original right prism and 𝟑𝟑 right triangular prisms that are each half of a cube. The 𝟑𝟑 right triangular prisms can be paired together to form 𝟑𝟑 cubes, or 𝟑𝟑 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟑𝟑. Altogether the area of the right triangular prism is (𝟏𝟏𝟏𝟏+ 𝟑𝟑) 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟑𝟑, or 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟑𝟑.

In both cases, slicing the square (or square face) along its diagonal divided the area of the square into two equal-sized triangular regions. When we sliced the right prism, however, what remained constant? The height of the given right rectangular prism and the resulting triangular prism are unchanged at

1 unit.

The argument used here is true in general for all right prisms. Since polygonal regions can be decomposed into triangles and rectangles, it is true that the polygonal base of a given right prism can be decomposed into triangular and rectangular regions that are bases of a set of right prisms that have heights equal to the height of the given right prism.

How could we create a right triangular prism with five times the volume of the right triangular prism pictured to the right, without changing the base? Draw your solution on the diagram, give the volume of the solid, and explain why your solution has five times the volume of the triangular prism.

If we stack five exact copies of the base (or bottom floor), the prism then has five times the number of unit cubes as the original, which means it has five times the volume, or 𝟗𝟗𝟗𝟗 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟑𝟑.

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What could we do to cut the volume of the right triangular prism pictured on the right in half without changing the base? Draw your solution on the diagram, give the volume of the solid, and explain why your solution has half the volume of the given triangular prism.

If we slice the height of the prism in half, each of the unit cubes that make up the triangular prism will have half the volume as in the original right triangular prism. The volume of the new right triangular prism is 𝟗𝟗 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝟑𝟑.

What can we conclude about how to find the volume of any right prism?

The volume of any right prism can be found by multiplying the area of its base times the height of the prism.

If we let 𝑉𝑉 represent the volume of a given right prism, let 𝐵𝐵 represent the area of the base of that given right prism, and let ℎ represent the height of that given right prism, then

𝑉𝑉 = 𝐵𝐵ℎ.

Have students complete the sentence below in their student materials.

To find the volume (𝑽𝑽) of any right prism …

Multiply the area of the right prism’s base (𝑩𝑩) times the height of the right prism (𝒉𝒉), 𝑽𝑽 = 𝑩𝑩𝒉𝒉.

Example (5 minutes): The Volume of a Right Triangular Prism

Students calculate the volume of a triangular prism that has not been decomposed from a rectangle.

Example: The Volume of a Right Triangular Prism

Find the volume of the right triangular prism shown in the diagram using 𝑽𝑽 = 𝑩𝑩𝒉𝒉.

𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑽𝑽 = �𝟏𝟏𝟐𝟐 𝒍𝒍𝒍𝒍�𝒉𝒉

𝑽𝑽 = �𝟏𝟏𝟐𝟐∙ 𝟒𝟒 𝐦𝐦 ∙

𝟏𝟏𝟐𝟐𝐦𝐦� ∙ 𝟑𝟑

𝟏𝟏𝟐𝟐

𝐦𝐦

𝑽𝑽 = �𝟐𝟐 𝐦𝐦 ∙𝟏𝟏𝟐𝟐

𝐦𝐦� ∙ 𝟑𝟑𝟏𝟏𝟐𝟐

𝐦𝐦

𝑽𝑽 = 𝟏𝟏 𝐦𝐦𝟐𝟐 ∙ 𝟑𝟑𝟏𝟏𝟐𝟐

𝐦𝐦

𝑽𝑽 = 𝟑𝟑𝟏𝟏𝟐𝟐 𝐦𝐦𝟑𝟑 The volume of the triangular prism is 𝟑𝟑𝟏𝟏𝟐𝟐 𝐦𝐦𝟑𝟑.

Scaffolding: Students often form the misconception that changing the dimensions of a given right prism will affect the prism’s volume by the same factor. Use this exercise to show that the volume of the cube is cut in half because the height is cut in half. If all dimensions of a unit cube were cut in half, the resulting volume would be

12∙ 12∙ 12

=18 , which is not

equal to 12

unit3.

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7•3 Lesson 23


Exercise (10 minutes): Multiple Volume Representations

Students find the volume of the right pentagonal prism using two different strategies.

Exercise: Multiple Volume Representations

The right pentagonal prism is composed of a right rectangular prism joined with a right triangular prism. Find the volume of the right pentagonal prism shown in the diagram using two different strategies.

Strategy #1

The volume of the pentagonal prism is equal to the sum of the volumes of the rectangular and triangular prisms.

𝑽𝑽 = 𝑽𝑽𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 𝐩𝐩𝐫𝐫𝐮𝐮𝐮𝐮𝐦𝐦 + 𝑽𝑽𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 𝐩𝐩𝐫𝐫𝐮𝐮𝐮𝐮𝐦𝐦

𝑽𝑽 = 𝑩𝑩𝒉𝒉 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑽𝑽 = (𝒍𝒍𝒍𝒍)𝒉𝒉 𝑽𝑽 = �𝟏𝟏𝟐𝟐 𝒍𝒍𝒍𝒍�𝒉𝒉

𝑽𝑽 = �𝟒𝟒 𝐦𝐦 ∙ 𝟑𝟑𝟏𝟏𝟐𝟐 𝐦𝐦� ∙ 𝟑𝟑𝟏𝟏𝟐𝟐 𝐦𝐦 𝑽𝑽 = �𝟏𝟏𝟐𝟐 ∙ 𝟒𝟒 𝐦𝐦 ∙ 𝟏𝟏𝟐𝟐 𝐦𝐦� ∙ 𝟑𝟑𝟏𝟏𝟐𝟐 𝐦𝐦

𝑽𝑽 = (𝟐𝟐𝟒𝟒 𝐦𝐦𝟐𝟐 + 𝟐𝟐 𝐦𝐦𝟐𝟐) ∙ 𝟑𝟑 𝟏𝟏𝟐𝟐 𝐦𝐦 𝑽𝑽 = �𝟐𝟐 𝐦𝐦 ∙ 𝟏𝟏𝟐𝟐 𝐦𝐦� ∙ 𝟑𝟑𝟏𝟏𝟐𝟐 𝐦𝐦

𝑽𝑽 = 𝟐𝟐𝟑𝟑 𝐦𝐦𝟐𝟐 ∙ 𝟑𝟑𝟏𝟏𝟐𝟐𝐦𝐦 𝑽𝑽 = (𝟏𝟏 𝐦𝐦𝟐𝟐) ∙ 𝟑𝟑 𝟏𝟏𝟐𝟐 𝐦𝐦

𝑽𝑽 = 𝟏𝟏𝟏𝟏𝟑𝟑 𝐦𝐦𝟑𝟑 + 𝟏𝟏𝟑𝟑 𝐦𝐦𝟑𝟑 𝑽𝑽 = 𝟑𝟑𝟏𝟏𝟐𝟐 𝐦𝐦𝟑𝟑

𝑽𝑽 = 𝟏𝟏𝟑𝟑𝟗𝟗 𝐦𝐦𝟑𝟑

So the total volume of the pentagonal prism is 𝟏𝟏𝟑𝟑𝟗𝟗 𝐦𝐦𝟑𝟑 + 𝟑𝟑𝟏𝟏𝟐𝟐 𝐦𝐦𝟑𝟑, or 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐 𝐦𝐦3.

Strategy #2

The volume of a right prism is equal to the area of its base times its height. The base is a rectangle and a triangle.


𝑩𝑩 = 𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 + 𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟒𝟒 𝐦𝐦 ∙ 𝟑𝟑𝟏𝟏𝟐𝟐 𝐦𝐦 𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟏𝟏𝟐𝟐 ∙ 𝟒𝟒 𝐦𝐦 ∙ 𝟏𝟏𝟐𝟐 𝐦𝐦 𝑽𝑽 = 𝟐𝟐𝟏𝟏 𝐦𝐦𝟐𝟐 ∙ 𝟑𝟑𝟏𝟏𝟐𝟐𝐦𝐦

𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟐𝟐𝟒𝟒 𝐦𝐦𝟐𝟐 + 𝟐𝟐 𝐦𝐦𝟐𝟐 𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟐𝟐 𝐦𝐦 ∙ 𝟏𝟏𝟐𝟐𝐦𝐦 𝑽𝑽 = 𝟏𝟏𝟑𝟑𝟐𝟐 𝐦𝐦𝟑𝟑 + 𝟏𝟏𝟑𝟑𝟏𝟏𝟐𝟐𝐦𝐦𝟑𝟑

𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟐𝟐𝟑𝟑 𝐦𝐦𝟐𝟐 𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟏𝟏 𝐦𝐦𝟐𝟐 𝑽𝑽 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐 𝐦𝐦𝟑𝟑

𝑩𝑩 = 𝟐𝟐𝟑𝟑 𝐦𝐦𝟐𝟐 + 𝟏𝟏 𝐦𝐦𝟐𝟐

𝑩𝑩 = 𝟐𝟐𝟏𝟏 𝐦𝐦𝟐𝟐

The volume of the right pentagonal prism is 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐𝐦𝐦𝟑𝟑.

Scaffolding: An alternative method that helps students visualize the connection between the area of the base, the height, and the volume of the right prism is to create pentagonal “floors” or “layers” with a depth of 1 unit. Students can physically pile the “floors” to form the right pentagonal prism. This example involves a fractional height so representation or visualization of a “floor” with a

height of 12 unit is necessary.

See below.

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Closing (2 minutes)

What are some strategies that we can use to find the volume of three-dimensional objects?

Find the area of the base, then multiply times the prism’s height; decompose the prism into two or more smaller prisms of the same height, and add the volumes of those smaller prisms.

The volume of a solid is always greater than or equal to zero.

If two solids are identical, they have equal volumes.

If a solid 𝑆𝑆 is the union of two non-overlapping solids 𝐴𝐴 and 𝐵𝐵, then the volume of solid 𝑆𝑆 is equal to the sum of the volumes of solids 𝐴𝐴 and 𝐵𝐵.


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Name Date


Exit Ticket The base of the right prism is a hexagon composed of a rectangle and two triangles. Find the volume of the right hexagonal prism using the formula 𝑉𝑉 = 𝐵𝐵ℎ.

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7•3 Lesson 23


𝟑𝟑𝟒𝟒

𝐮𝐮𝐮𝐮.

𝟑𝟑𝟒𝟒

𝐮𝐮𝐮𝐮.

𝟑𝟑𝟒𝟒

𝐮𝐮𝐮𝐮.


The base of the right prism is a hexagon composed of a rectangle and two triangles. Find the volume of the right hexagonal prism using the formula 𝑽𝑽 = 𝑩𝑩𝒉𝒉.

The area of the base is the sum of the areas of the rectangle and the two triangles.

𝑩𝑩 = 𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 + 𝟐𝟐 ∙ 𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫

𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝒍𝒍𝒍𝒍 𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟏𝟏𝟐𝟐 𝒍𝒍𝒍𝒍

𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟐𝟐𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮. ∙ 𝟏𝟏 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. 𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟏𝟏𝟐𝟐 �𝟏𝟏

𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. ∙ 𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮. �

𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = �𝟗𝟗𝟒𝟒 ⋅𝟑𝟑𝟐𝟐� 𝐮𝐮𝐮𝐮𝟐𝟐 𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = �𝟏𝟏𝟐𝟐 ⋅

𝟑𝟑𝟐𝟐 ⋅

𝟑𝟑𝟒𝟒� 𝐮𝐮𝐮𝐮

𝟐𝟐

𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟐𝟐𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟗𝟗

𝟏𝟏𝟑𝟑 𝐮𝐮𝐮𝐮𝟐𝟐

𝑩𝑩 = 𝟐𝟐𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟐𝟐� 𝟗𝟗𝟏𝟏𝟑𝟑 𝐮𝐮𝐮𝐮𝟐𝟐� 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = 𝟐𝟐𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟗𝟗

𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 𝑽𝑽 = �𝟗𝟗𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐� ∙ 𝟑𝟑 𝐮𝐮𝐮𝐮.

𝑩𝑩 = 𝟑𝟑𝟑𝟑𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 𝑽𝑽 = 𝟐𝟐𝟏𝟏

𝟐𝟐 𝐮𝐮𝐮𝐮𝟑𝟑

𝑩𝑩 = 𝟗𝟗𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 𝑽𝑽 = 𝟏𝟏𝟑𝟑𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮

𝟑𝟑

The volume of the hexagonal prism is 𝟏𝟏𝟑𝟑𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟑𝟑.


1. Calculate the volume of each solid using the formula 𝑽𝑽 = 𝑩𝑩𝒉𝒉 (all angles are 𝟗𝟗𝟗𝟗 degrees).

a. 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑽𝑽 = (𝟏𝟏 𝐫𝐫𝐦𝐦 ∙ 𝟏𝟏 𝐫𝐫𝐦𝐦) ∙ 𝟏𝟏𝟐𝟐 𝟏𝟏𝟐𝟐𝐫𝐫𝐦𝐦

𝑽𝑽 = �𝟏𝟏𝟑𝟑 ∙ 𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐� 𝐫𝐫𝐦𝐦𝟑𝟑

𝑽𝑽 = 𝟑𝟑𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟑𝟑 + 𝟐𝟐𝟏𝟏 𝐫𝐫𝐦𝐦𝟑𝟑

𝑽𝑽 = 𝟏𝟏𝟗𝟗𝟗𝟗 𝐫𝐫𝐦𝐦𝟑𝟑

The volume of the solid is 𝟏𝟏𝟗𝟗𝟗𝟗 𝐫𝐫𝐦𝐦𝟑𝟑.

b. 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑽𝑽 = �𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮. ∙ 𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮. � ∙ 𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮.

𝑽𝑽 = � 𝟗𝟗𝟏𝟏𝟑𝟑� ∙𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮

𝟑𝟑

𝑽𝑽 = 𝟐𝟐𝟏𝟏𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮𝟑𝟑

The volume of the cube is 𝟐𝟐𝟏𝟏𝟑𝟑𝟒𝟒

𝐮𝐮𝐮𝐮𝟑𝟑.

𝟏𝟏 𝐫𝐫𝐦𝐦

𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐

𝐫𝐫𝐦𝐦

𝟏𝟏 𝐫𝐫𝐦𝐦

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c. 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = 𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 + 𝑨𝑨𝐮𝐮𝐬𝐬𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫

𝑩𝑩 = 𝒍𝒍𝒍𝒍 + 𝒔𝒔𝟐𝟐

𝑩𝑩 = �𝟐𝟐𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮.∙ 𝟒𝟒 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. � + �𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. �𝟐𝟐

𝑩𝑩 = �𝟏𝟏𝟗𝟗 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐� + �𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. ∙ 𝟏𝟏 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. � 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐 + �𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐� 𝑽𝑽 = 𝟏𝟏𝟑𝟑𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 ∙ 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮.

𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 𝑽𝑽 = 𝟏𝟏𝟑𝟑

𝟐𝟐 𝐮𝐮𝐮𝐮𝟑𝟑 + 𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮𝟑𝟑

𝑩𝑩 = 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 𝑽𝑽 = 𝟑𝟑 𝐮𝐮𝐮𝐮𝟑𝟑 + 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟑𝟑 + 𝟏𝟏

𝟒𝟒 𝐮𝐮𝐮𝐮𝟑𝟑

𝑩𝑩 = 𝟏𝟏𝟑𝟑𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 𝑽𝑽 = 𝟑𝟑𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮𝟑𝟑

The volume of the solid is 𝟑𝟑𝟑𝟑𝟒𝟒 𝐮𝐮𝐮𝐮𝟑𝟑.

d. 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = (𝑨𝑨𝐫𝐫𝐫𝐫 𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫) − (𝑨𝑨𝐮𝐮𝐦𝐦 𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫)

𝑩𝑩 = (𝒍𝒍𝒍𝒍)𝟏𝟏 − (𝒍𝒍𝒍𝒍)𝟐𝟐

𝑩𝑩 = (𝟑𝟑 𝐲𝐲𝐲𝐲. ∙ 𝟒𝟒 𝐲𝐲𝐲𝐲. ) − �𝟏𝟏𝟏𝟏𝟑𝟑 𝐲𝐲𝐲𝐲. ∙ 𝟐𝟐 𝐲𝐲𝐲𝐲. � 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = 𝟐𝟐𝟒𝟒 𝐲𝐲𝐲𝐲𝟐𝟐 − �𝟐𝟐 𝐲𝐲𝐲𝐲𝟐𝟐 + 𝟐𝟐𝟑𝟑 𝐲𝐲𝐲𝐲𝟐𝟐� 𝑽𝑽 = �𝟐𝟐𝟏𝟏𝟏𝟏𝟑𝟑 𝐲𝐲𝐲𝐲𝟐𝟐� ∙ 𝟐𝟐𝟑𝟑 𝐲𝐲𝐲𝐲.

𝑩𝑩 = 𝟐𝟐𝟒𝟒 𝐲𝐲𝐲𝐲𝟐𝟐 − 𝟐𝟐 𝐲𝐲𝐲𝐲𝟐𝟐 − 𝟐𝟐𝟑𝟑 𝐲𝐲𝐲𝐲𝟐𝟐 𝑽𝑽 = 𝟏𝟏𝟒𝟒 𝐲𝐲𝐲𝐲𝟑𝟑 + �𝟏𝟏𝟑𝟑 𝐲𝐲𝐲𝐲𝟐𝟐 ∙ 𝟐𝟐𝟑𝟑 𝐲𝐲𝐲𝐲. �

𝑩𝑩 = 𝟐𝟐𝟐𝟐 𝐲𝐲𝐲𝐲𝟐𝟐 − 𝟐𝟐𝟑𝟑 𝐲𝐲𝐲𝐲𝟐𝟐 𝑽𝑽 = 𝟏𝟏𝟒𝟒 𝐲𝐲𝐲𝐲𝟑𝟑 + 𝟐𝟐

𝟗𝟗 𝐲𝐲𝐲𝐲𝟑𝟑

𝑩𝑩 = 𝟐𝟐𝟏𝟏𝟏𝟏𝟑𝟑 𝐲𝐲𝐲𝐲𝟐𝟐 𝑽𝑽 = 𝟏𝟏𝟒𝟒𝟐𝟐𝟗𝟗 𝐲𝐲𝐲𝐲𝟑𝟑

The volume of the solid is 𝟏𝟏𝟒𝟒𝟐𝟐𝟗𝟗 𝐲𝐲𝐲𝐲𝟑𝟑.

e. 𝑽𝑽 = 𝑩𝑩𝒉𝒉𝐩𝐩𝐫𝐫𝐮𝐮𝐮𝐮𝐦𝐦

𝑩𝑩 = 𝟏𝟏𝟐𝟐𝒃𝒃𝒉𝒉𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = 𝟏𝟏𝟐𝟐 ∙ 𝟒𝟒 𝐫𝐫𝐦𝐦 ∙ 𝟒𝟒 𝐫𝐫𝐦𝐦 𝑽𝑽 = 𝟏𝟏 𝐫𝐫𝐦𝐦𝟐𝟐 ∙ 𝟑𝟑 𝟏𝟏

𝟏𝟏𝟗𝟗 𝐫𝐫𝐦𝐦

𝑩𝑩 = 𝟐𝟐 ∙ 𝟒𝟒 𝐫𝐫𝐦𝐦𝟐𝟐 𝑽𝑽 = 𝟒𝟒𝟏𝟏 𝐫𝐫𝐦𝐦𝟑𝟑 + 𝟏𝟏𝟑𝟑𝟏𝟏𝟗𝟗 𝐫𝐫𝐦𝐦𝟑𝟑

𝑩𝑩 = 𝟏𝟏 𝐫𝐫𝐦𝐦𝟐𝟐 𝑽𝑽 = 𝟒𝟒𝟏𝟏 𝐫𝐫𝐦𝐦𝟑𝟑 + 𝟏𝟏 𝐫𝐫𝐦𝐦𝟑𝟑 + 𝟑𝟑𝟏𝟏𝟗𝟗 𝐫𝐫𝐦𝐦𝟑𝟑

𝑽𝑽 = 𝟏𝟏𝟑𝟑 𝐫𝐫𝐦𝐦𝟑𝟑 + 𝟑𝟑𝟏𝟏 𝐫𝐫𝐦𝐦𝟑𝟑

𝑽𝑽 = 𝟏𝟏𝟑𝟑𝟑𝟑𝟏𝟏 𝐫𝐫𝐦𝐦𝟑𝟑

The volume of the solid is 𝟏𝟏𝟑𝟑𝟑𝟑𝟏𝟏 𝐫𝐫𝐦𝐦𝟑𝟑.

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f. 𝑽𝑽 = 𝑩𝑩𝒉𝒉𝒑𝒑𝒑𝒑𝒑𝒑𝒔𝒔𝒑𝒑

𝑩𝑩 = 𝟏𝟏𝟐𝟐𝒃𝒃𝒉𝒉𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = 𝟏𝟏𝟐𝟐 ∙ 𝟗𝟗

𝟑𝟑𝟐𝟐𝟏𝟏 𝐮𝐮𝐮𝐮. ∙ 𝟐𝟐 𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. 𝑽𝑽 = �𝟏𝟏𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐� ∙ 𝟏𝟏 𝐮𝐮𝐮𝐮.

𝑩𝑩 = 𝟏𝟏𝟐𝟐 ∙ 𝟐𝟐

𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. ∙ 𝟗𝟗 𝟑𝟑

𝟐𝟐𝟏𝟏 𝐮𝐮𝐮𝐮. 𝑽𝑽 = 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟑𝟑

𝑩𝑩 = �𝟏𝟏𝟏𝟏𝟒𝟒� ∙ �𝟗𝟗𝟑𝟑𝟐𝟐𝟏𝟏� 𝐮𝐮𝐮𝐮

𝟐𝟐

𝑩𝑩 = �𝟏𝟏𝟒𝟒 ⋅𝟐𝟐𝟐𝟐𝟏𝟏𝟐𝟐𝟏𝟏 � 𝐮𝐮𝐮𝐮

𝟐𝟐

𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 The volume of the solid is 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟑𝟑.

g. 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = 𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 + 𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = 𝒍𝒍𝒍𝒍 + 𝟏𝟏𝟐𝟐𝒃𝒃𝒉𝒉 𝑽𝑽 = 𝟐𝟐𝟑𝟑𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟐𝟐 ∙ 𝟗𝟗 𝐫𝐫𝐦𝐦

𝑩𝑩 = �𝟏𝟏𝟏𝟏𝟒𝟒 𝐫𝐫𝐦𝐦 ∙ 𝟒𝟒 𝐫𝐫𝐦𝐦� + 𝟏𝟏𝟐𝟐�𝟒𝟒 𝐫𝐫𝐦𝐦 ∙ 𝟏𝟏𝟏𝟏𝟒𝟒 𝐫𝐫𝐦𝐦� 𝑽𝑽 = 𝟐𝟐𝟗𝟗𝟏𝟏 𝐫𝐫𝐦𝐦𝟑𝟑 + 𝟗𝟗

𝟐𝟐 𝐫𝐫𝐦𝐦𝟑𝟑

𝑩𝑩 = (𝟐𝟐𝟗𝟗 𝐫𝐫𝐦𝐦𝟐𝟐 + 𝟏𝟏 𝐫𝐫𝐦𝐦𝟐𝟐) + �𝟐𝟐 𝐫𝐫𝐦𝐦 ∙ 𝟏𝟏𝟏𝟏𝟒𝟒 𝐫𝐫𝐦𝐦� 𝑽𝑽 = 𝟐𝟐𝟗𝟗𝟏𝟏 𝐫𝐫𝐦𝐦𝟑𝟑 + 𝟒𝟒 𝐫𝐫𝐦𝐦𝟑𝟑 + 𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟑𝟑

𝑩𝑩 = 𝟐𝟐𝟏𝟏 𝐫𝐫𝐦𝐦𝟐𝟐 + 𝟐𝟐 𝐫𝐫𝐦𝐦𝟐𝟐 + 𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟐𝟐 𝑽𝑽 = 𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟑𝟑

𝑩𝑩 = 𝟐𝟐𝟑𝟑 𝐫𝐫𝐦𝐦𝟐𝟐 + 𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟐𝟐

𝑩𝑩 = 𝟐𝟐𝟑𝟑𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟐𝟐 The volume of the solid is 𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟑𝟑.

h. 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = 𝑨𝑨𝐫𝐫𝐫𝐫𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 + 𝟐𝟐𝑨𝑨𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐮𝐮𝐫𝐫𝐫𝐫𝐫𝐫 𝑽𝑽 = 𝑩𝑩𝒉𝒉

𝑩𝑩 = 𝒍𝒍𝒍𝒍 + 𝟐𝟐 ∙ 𝟏𝟏𝟐𝟐𝒃𝒃𝒉𝒉 𝑽𝑽 = 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐 ∙ 𝟐𝟐 𝐮𝐮𝐮𝐮.

𝑩𝑩 = �𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮. ∙ 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮. � + �𝟏𝟏 ∙ 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮.∙ 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮. � 𝑽𝑽 = 𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮𝟑𝟑

𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟗𝟗 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏

𝟒𝟒𝟗𝟗 𝐮𝐮𝐮𝐮𝟐𝟐

𝑩𝑩 = 𝟒𝟒𝟒𝟒𝟗𝟗 𝐮𝐮𝐮𝐮𝟐𝟐 + 𝟏𝟏

𝟒𝟒𝟗𝟗 𝐮𝐮𝐮𝐮𝟐𝟐 The volume of the solid is 𝟏𝟏𝟒𝟒

𝐮𝐮𝐮𝐮𝟑𝟑.

𝑩𝑩 = 𝟏𝟏𝟒𝟒𝟗𝟗 𝐮𝐮𝐮𝐮𝟐𝟐

𝑩𝑩 = 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟐𝟐

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2. Let 𝒍𝒍 represent the length, 𝒍𝒍 the width, and 𝒉𝒉 the height of a right rectangular prism. Find the volume of the prism when

a. 𝒍𝒍 = 𝟑𝟑 𝐫𝐫𝐦𝐦, 𝒍𝒍 = 𝟐𝟐𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦, and 𝒉𝒉 = 𝟏𝟏 𝐫𝐫𝐦𝐦.

𝑽𝑽 = 𝒍𝒍𝒍𝒍𝒉𝒉

𝑽𝑽 = 𝟑𝟑 𝐫𝐫𝐦𝐦 ∙ 𝟐𝟐𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦 ∙ 𝟏𝟏 𝐫𝐫𝐦𝐦

𝑽𝑽 = 𝟐𝟐𝟏𝟏 ∙ �𝟐𝟐𝟏𝟏𝟐𝟐� 𝐫𝐫𝐦𝐦𝟑𝟑

𝑽𝑽 = 𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟑𝟑 The volume of the prism is 𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟑𝟑.

b. 𝒍𝒍 = 𝟏𝟏𝟒𝟒 𝐫𝐫𝐦𝐦, 𝒍𝒍 = 𝟒𝟒 𝐫𝐫𝐦𝐦, and 𝒉𝒉 = 𝟏𝟏𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦.

𝑽𝑽 = 𝒍𝒍𝒍𝒍𝒉𝒉

𝑽𝑽 = 𝟏𝟏𝟒𝟒 𝐫𝐫𝐦𝐦 ∙ 𝟒𝟒 𝐫𝐫𝐦𝐦 ∙ 𝟏𝟏𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦

𝑽𝑽 = 𝟏𝟏𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟑𝟑 The volume of the prism is 𝟏𝟏𝟏𝟏𝟐𝟐 𝐫𝐫𝐦𝐦𝟑𝟑.

3. Find the length of the edge indicated in each diagram.

a. 𝑽𝑽 = 𝑩𝑩𝒉𝒉 Let 𝒉𝒉 represent the number of inches in the height of the prism.

𝟗𝟗𝟑𝟑𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟑𝟑 = 𝟐𝟐𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐 ∙ 𝒉𝒉

𝟗𝟗𝟑𝟑𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮𝟑𝟑 = 𝟐𝟐𝟐𝟐𝒉𝒉 𝐮𝐮𝐮𝐮𝟐𝟐

𝟐𝟐𝟐𝟐𝒉𝒉 = 𝟗𝟗𝟑𝟑.𝟏𝟏 𝐮𝐮𝐮𝐮

𝒉𝒉 = 𝟒𝟒.𝟐𝟐𝟏𝟏 𝐮𝐮𝐮𝐮

The height of the right rectangular prism is 𝟒𝟒𝟏𝟏𝟒𝟒 𝐮𝐮𝐮𝐮.

What are possible dimensions of the base?

𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮 by 𝟐𝟐 𝐮𝐮𝐮𝐮, or 𝟐𝟐𝟐𝟐 𝐮𝐮𝐮𝐮 by 𝟏𝟏 𝐮𝐮𝐮𝐮

b. 𝑽𝑽 = 𝑩𝑩𝒉𝒉 Let 𝒉𝒉 represent the number of meters in the height of the triangular base of the prism.

𝑽𝑽 = �𝟏𝟏𝟐𝟐𝒃𝒃𝒉𝒉𝒕𝒕𝒑𝒑𝒑𝒑𝒕𝒕𝒕𝒕𝒕𝒕𝒍𝒍𝒕𝒕� ∙ 𝒉𝒉𝒑𝒑𝒑𝒑𝒑𝒑𝒔𝒔𝒑𝒑

𝟒𝟒𝟏𝟏𝟐𝟐 𝐦𝐦𝟑𝟑 = �𝟏𝟏𝟐𝟐 ∙ 𝟑𝟑 𝐦𝐦 ∙ 𝒉𝒉� ∙ 𝟑𝟑 𝐦𝐦

𝟒𝟒𝟏𝟏𝟐𝟐 𝐦𝐦𝟑𝟑 = 𝟏𝟏𝟐𝟐 ∙ 𝟏𝟏𝟏𝟏 𝐦𝐦𝟐𝟐 ∙ 𝒉𝒉

𝟒𝟒𝟏𝟏𝟐𝟐 𝐦𝐦𝟑𝟑 = 𝟗𝟗𝒉𝒉 𝐦𝐦𝟐𝟐

𝟗𝟗𝒉𝒉 = 𝟒𝟒.𝟏𝟏 𝐦𝐦

𝒉𝒉 = 𝟗𝟗.𝟏𝟏 𝐦𝐦

The height of the triangle is 𝟏𝟏𝟐𝟐𝐦𝐦.

𝐀𝐀𝐫𝐫𝐫𝐫𝐫𝐫 = 𝟐𝟐𝟐𝟐 𝐮𝐮𝐮𝐮𝟐𝟐

?

𝐕𝐕𝐕𝐕𝐫𝐫𝐮𝐮𝐦𝐦𝐫𝐫 = 𝟗𝟗𝟑𝟑𝟏𝟏𝟐𝟐

𝐮𝐮𝐮𝐮𝟑𝟑

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4. The volume of a cube is 𝟑𝟑𝟑𝟑𝟏𝟏 𝐮𝐮𝐮𝐮𝟑𝟑. Find the length of each edge of the cube.

𝑽𝑽 = 𝒔𝒔𝟑𝟑, and since the volume is a fraction, the edge length must also be fractional.

𝟑𝟑𝟑𝟑𝟏𝟏 𝐮𝐮𝐮𝐮𝟑𝟑 = 𝟐𝟐𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝟑𝟑

𝟑𝟑𝟑𝟑𝟏𝟏 𝐮𝐮𝐮𝐮𝟑𝟑 = 𝟑𝟑𝟐𝟐 𝐮𝐮𝐮𝐮. ∙ 𝟑𝟑𝟐𝟐 𝐮𝐮𝐮𝐮. ∙ 𝟑𝟑𝟐𝟐 𝐮𝐮𝐮𝐮.

𝟑𝟑𝟑𝟑𝟏𝟏 𝐮𝐮𝐮𝐮𝟑𝟑 = �𝟑𝟑𝟐𝟐 𝐮𝐮𝐮𝐮. �𝟑𝟑

The lengths of the edges of the cube are 𝟑𝟑𝟐𝟐 𝐮𝐮𝐮𝐮., or 𝟏𝟏𝟏𝟏𝟐𝟐 𝐮𝐮𝐮𝐮.

5. Given a right rectangular prism with a volume of 𝟏𝟏𝟏𝟏𝟐𝟐 𝐟𝐟𝐮𝐮𝟑𝟑, a length of 𝟏𝟏 𝐟𝐟𝐮𝐮., and a width of 𝟐𝟐 𝐟𝐟𝐮𝐮., find the height of the prism.


𝑽𝑽 = (𝒍𝒍𝒍𝒍)𝒉𝒉 Let 𝒉𝒉 represent the number of feet in the height of the prism.

𝟏𝟏𝟏𝟏𝟐𝟐

𝐟𝐟𝐮𝐮𝟑𝟑 = (𝟏𝟏𝐟𝐟𝐮𝐮. ∙ 𝟐𝟐𝐟𝐟𝐮𝐮. ) ∙ 𝒉𝒉

𝟏𝟏𝟏𝟏𝟐𝟐

𝐟𝐟𝐮𝐮𝟑𝟑 = 𝟏𝟏𝟗𝟗 𝐟𝐟𝐮𝐮𝟐𝟐 ∙ 𝒉𝒉

𝟏𝟏.𝟏𝟏 𝐟𝐟𝐮𝐮𝟑𝟑 = 𝟏𝟏𝟗𝟗𝒉𝒉 𝐟𝐟𝐮𝐮𝟐𝟐 𝒉𝒉 = 𝟗𝟗.𝟏𝟏𝟏𝟏 𝐟𝐟𝐮𝐮.

The height of the right rectangular prism is 𝟑𝟑𝟒𝟒 𝐟𝐟𝐮𝐮. (or 𝟗𝟗 𝐮𝐮𝐮𝐮.).

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Student Outcomes

Students use the formula for the volume of a right rectangular prism to answer questions about the capacity of tanks.

Students compute volumes of right prisms involving fractional values for length.

Lesson Notes Students extend their knowledge about the volume of solid figures to the notion of liquid volume. The Opening Exercise requires a small amount of water. Have an absorbent towel available to soak up the water at the completion of the exercise.

Classwork


Pour enough water onto a large flat surface to form a puddle. Have students discuss how to determine the volume of the water. Provide 2 minutes for student discussion, and then start the class discussion.


Why can’t we easily determine the volume of the water in the puddle?

The puddle does not have any definite shape or depth that we can easily measure. How can we measure the volume of the water in three dimensions?

The volume can be measured in three dimensions if put into a container. In a container, such as a prism, water takes on the shape of the container. We can measure the dimensions of the container to determine an approximate volume of the water in cubic units.

Exploratory Challenge (8 minutes): Measuring a Container’s Capacity

Students progress from measuring the volume of a liquid inside a right rectangular prism filled to capacity to solving a variety of problems involving liquids and prism-shaped containers.

Ask questions to guide students in discovering the need to account for the thickness of the container material in determining the “inside” volume of the container. For instance, ask, “Is the length of the inside of the container 12 inches? Why not? What is the width of the inside container? The depth? Why did you have to subtract twice the thickness to get the length and width, but only one times the thickness to get the depth?

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Exploratory Challenge: Measuring a Container’s Capacity

A box in the shape of a right rectangular prism has a length of 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢, a width of 𝟔𝟔 𝐢𝐢𝐢𝐢., and a height of 𝟖𝟖 𝐢𝐢𝐢𝐢. The base and

the walls of the container are 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢 thick, and its top is open. What is the capacity of the right rectangular prism? (Hint: The capacity is equal to the volume of water needed to fill the prism to the top.)

If the prism is filled with water, the water will take the shape of a right rectangular prism slightly smaller than the

container. The dimensions of the smaller prism are a length of 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢, a width of 𝟓𝟓𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢, and a height of 𝟕𝟕𝟑𝟑𝟒𝟒 𝐢𝐢𝐢𝐢.

𝑽𝑽 = 𝑩𝑩𝑩𝑩

𝑽𝑽 = (𝒍𝒍𝒍𝒍)𝑩𝑩

𝑽𝑽 = �𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 ∙ 𝟓𝟓 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢� ∙ 𝟕𝟕 𝟑𝟑𝟒𝟒 𝐢𝐢𝐢𝐢

𝑽𝑽 = �𝟏𝟏𝟑𝟑𝟏𝟏

𝐢𝐢𝐢𝐢 ∙𝟏𝟏𝟏𝟏𝟏𝟏

𝐢𝐢𝐢𝐢� ∙𝟑𝟑𝟏𝟏𝟒𝟒

𝐢𝐢𝐢𝐢

𝑽𝑽 = �𝟏𝟏𝟓𝟓𝟑𝟑𝟒𝟒

𝐢𝐢𝐢𝐢𝟏𝟏� ∙𝟑𝟑𝟏𝟏𝟒𝟒

𝐢𝐢𝐢𝐢

𝑽𝑽 =𝟕𝟕𝟖𝟖𝟒𝟒𝟑𝟑𝟏𝟏𝟔𝟔

𝐢𝐢𝐢𝐢𝟑𝟑

𝑽𝑽 = 𝟒𝟒𝟒𝟒𝟒𝟒 𝟑𝟑𝟏𝟏𝟔𝟔 𝐢𝐢𝐢𝐢𝟑𝟑

The capacity of the right rectangular prism is 𝟒𝟒𝟒𝟒𝟒𝟒 𝟑𝟑𝟏𝟏𝟔𝟔 𝐢𝐢𝐢𝐢𝟑𝟑.

Example 1 (5 minutes): Measuring Liquid in a Container in Three Dimensions

Students use the inside of right prism-shaped containers to calculate the volumes of contained liquids.

Example 1: Measuring Liquid in a Container in Three Dimensions

A glass container is in the form of a right rectangular prism. The container is 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜 long, 𝟖𝟖 𝐜𝐜𝐜𝐜 wide, and 𝟑𝟑𝟒𝟒 𝐜𝐜𝐜𝐜 high. The top of the container is open, and the base and walls of the container are 𝟑𝟑 𝐜𝐜𝐜𝐜 (or 𝟒𝟒.𝟑𝟑 𝐜𝐜𝐜𝐜) thick. The water in the container is 𝟔𝟔 𝐜𝐜𝐜𝐜 from the top of the container. What is the volume of the water in the container?

Because of the walls and base of the container, the water in the container forms a right rectangular prism that is 𝟒𝟒.𝟒𝟒 𝐜𝐜𝐜𝐜 long, 𝟕𝟕.𝟒𝟒 𝐜𝐜𝐜𝐜 wide, and 𝟏𝟏𝟑𝟑.𝟕𝟕 𝐜𝐜𝐜𝐜 tall.



𝑽𝑽 = (𝟒𝟒.𝟒𝟒 𝐜𝐜𝐜𝐜 ∙ 𝟕𝟕.𝟒𝟒 𝐜𝐜𝐜𝐜) ∙ 𝟏𝟏𝟑𝟑.𝟕𝟕 𝐜𝐜𝐜𝐜

𝑽𝑽 = �𝟒𝟒𝟒𝟒𝟏𝟏𝟒𝟒

𝐜𝐜𝐜𝐜 ∙𝟕𝟕𝟒𝟒𝟏𝟏𝟒𝟒

𝐜𝐜𝐜𝐜� ∙𝟏𝟏𝟑𝟑𝟕𝟕𝟏𝟏𝟒𝟒

𝐜𝐜𝐜𝐜

𝑽𝑽 = �𝟔𝟔,𝟒𝟒𝟓𝟓𝟔𝟔𝟏𝟏𝟒𝟒𝟒𝟒

𝐜𝐜𝐜𝐜𝟏𝟏� ∙𝟏𝟏𝟑𝟑𝟕𝟕𝟏𝟏𝟒𝟒

𝐜𝐜𝐜𝐜

𝑽𝑽 =𝟏𝟏,𝟔𝟔𝟒𝟒𝟖𝟖,𝟓𝟓𝟕𝟕𝟏𝟏𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒

𝐜𝐜𝐜𝐜𝟑𝟑

𝑽𝑽 = 𝟏𝟏,𝟔𝟔𝟒𝟒𝟖𝟖.𝟓𝟓𝟕𝟕𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑

The volume of the water in the container is 𝟏𝟏,𝟔𝟔𝟒𝟒𝟖𝟖.𝟔𝟔 𝐜𝐜𝐜𝐜𝟑𝟑.

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Students determine the depth of a given volume of water in a container of a given size.

Example 2

𝟕𝟕.𝟏𝟏 𝐋𝐋 of water are poured into a container in the shape of a right rectangular prism. The inside of the container is 𝟓𝟓𝟒𝟒 𝐜𝐜𝐜𝐜 long, 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜 wide, and 𝟏𝟏𝟓𝟓 𝐜𝐜𝐜𝐜 tall. How far from the top of the container is the surface of the water? (𝟏𝟏 𝐋𝐋 = 𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑)

𝟕𝟕.𝟏𝟏 𝐋𝐋 = 𝟕𝟕,𝟏𝟏𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑

𝑽𝑽 = 𝑩𝑩𝑩𝑩 𝑽𝑽 = (𝒍𝒍𝒍𝒍)𝑩𝑩

𝟕𝟕,𝟏𝟏𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = (𝟓𝟓𝟒𝟒 𝐜𝐜𝐜𝐜)(𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜)𝑩𝑩 𝟕𝟕,𝟏𝟏𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟕𝟕,𝟏𝟏𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 ∙𝟏𝟏

𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 = 𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙𝟏𝟏

𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟕𝟕,𝟏𝟏𝟒𝟒𝟒𝟒𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒

𝐜𝐜𝐜𝐜 = 𝟏𝟏 ∙ 𝑩𝑩

𝟕𝟕.𝟏𝟏 𝐜𝐜𝐜𝐜 = 𝑩𝑩

The depth of the water is 𝟕𝟕.𝟏𝟏 𝐜𝐜𝐜𝐜. The height of the container is 𝟏𝟏𝟓𝟓 𝐜𝐜𝐜𝐜.

𝟏𝟏𝟓𝟓 𝐜𝐜𝐜𝐜− 𝟕𝟕.𝟏𝟏 𝐜𝐜𝐜𝐜 = 𝟏𝟏𝟕𝟕.𝟖𝟖 𝐜𝐜𝐜𝐜

The surface of the water is 𝟏𝟏𝟕𝟕.𝟖𝟖 𝐜𝐜𝐜𝐜 from the top of the container.


Students find unknown measurements of a right prism given its volume and two dimensions.

Example 3

A fuel tank is the shape of a right rectangular prism and has 𝟏𝟏𝟕𝟕 𝐋𝐋 of fuel in it. It is determined that the tank is 𝟑𝟑𝟒𝟒

full. The

inside dimensions of the base of the tank are 𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜 by 𝟓𝟓𝟒𝟒 𝐜𝐜𝐜𝐜. What is the height of the fuel in the tank? How deep is the tank? (𝟏𝟏 𝐋𝐋 = 𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑)

Let the height of the fuel in the tank be 𝑩𝑩 𝐜𝐜𝐜𝐜.

𝟏𝟏𝟕𝟕 𝐋𝐋 = 𝟏𝟏𝟕𝟕,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑


𝟏𝟏𝟕𝟕,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = (𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜 ∙ 𝟓𝟓𝟒𝟒 𝐜𝐜𝐜𝐜) ∙ 𝑩𝑩 𝟏𝟏𝟕𝟕,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = (𝟒𝟒,𝟓𝟓𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏) ∙ 𝑩𝑩

𝟏𝟏𝟕𝟕,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 ∙𝟏𝟏

𝟒𝟒,𝟓𝟓𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 = 𝟒𝟒,𝟓𝟓𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙𝟏𝟏

𝟒𝟒,𝟓𝟓𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟏𝟏𝟕𝟕,𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒,𝟓𝟓𝟒𝟒𝟒𝟒


𝟔𝟔 𝐜𝐜𝐜𝐜 = 𝑩𝑩

The height of the fuel in the tank is 𝟔𝟔 𝐜𝐜𝐜𝐜. The height of the fuel is 𝟑𝟑𝟒𝟒 the depth of the tank. Let 𝒅𝒅 represent the depth of the tank in centimeters.

𝟔𝟔 𝐜𝐜𝐜𝐜 =𝟑𝟑𝟒𝟒𝒅𝒅

𝟔𝟔 𝐜𝐜𝐜𝐜 ∙𝟒𝟒𝟑𝟑

=𝟑𝟑𝟒𝟒∙𝟒𝟒𝟑𝟑∙ 𝒅𝒅

𝟖𝟖 𝐜𝐜𝐜𝐜 = 𝒅𝒅 The depth of the fuel tank is 𝟖𝟖 𝐜𝐜𝐜𝐜.

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Closing (2 minutes)

How do containers, such as prisms, allow us to measure the volumes of liquids using three dimensions?

When liquid is poured into a container, the liquid takes on the shape of the container’s interior. We can measure the volume of prisms in three dimensions, allowing us to measure the volume of the liquid in three dimensions.

What special considerations have to be made when measuring liquids in containers in three dimensions?

The outside and inside dimensions of a container will not be the same because the container has wall thickness. In addition, whether or not the container is filled to capacity will affect the volume of the liquid in the container.


Students may be allowed to use calculators when completing this Exit Ticket.

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Name Date


Exit Ticket Lawrence poured 27.328 L of water into a right rectangular prism-shaped tank. The base of the tank is 40 cm by 28 cm.

When he finished pouring the water, the tank was 23 full. (1 L = 1,000 cm3)

a. How deep is the water in the tank?

b. How deep is the tank?

c. How many liters of water can the tank hold in total?

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Lawrence poured 𝟏𝟏𝟕𝟕.𝟑𝟑𝟏𝟏𝟖𝟖 𝐋𝐋 of water into a right rectangular prism-shaped tank. The base of the tank is 𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜 by

𝟏𝟏𝟖𝟖 𝐜𝐜𝐜𝐜. When he finished pouring the water, the tank was 𝟏𝟏𝟑𝟑 full. (𝟏𝟏 𝐋𝐋 = 𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑)

a. How deep is the water in the tank?

𝟏𝟏𝟕𝟕.𝟑𝟑𝟏𝟏𝟖𝟖 𝐋𝐋 = 𝟏𝟏𝟕𝟕,𝟑𝟑𝟏𝟏𝟖𝟖 𝐜𝐜𝐜𝐜𝟑𝟑


𝟏𝟏𝟕𝟕,𝟑𝟑𝟏𝟏𝟖𝟖 𝐜𝐜𝐜𝐜𝟑𝟑 = (𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏𝟖𝟖 𝐜𝐜𝐜𝐜) ∙ 𝑩𝑩 𝟏𝟏𝟕𝟕,𝟑𝟑𝟏𝟏𝟖𝟖 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟏𝟏,𝟏𝟏𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟏𝟏𝟕𝟕,𝟑𝟑𝟏𝟏𝟖𝟖 𝐜𝐜𝐜𝐜𝟑𝟑 ∙𝟏𝟏

𝟏𝟏,𝟏𝟏𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 = 𝟏𝟏,𝟏𝟏𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙𝟏𝟏

𝟏𝟏,𝟏𝟏𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟏𝟏𝟕𝟕,𝟑𝟑𝟏𝟏𝟖𝟖𝟏𝟏,𝟏𝟏𝟏𝟏𝟒𝟒


𝟏𝟏𝟒𝟒𝟏𝟏𝟖𝟖𝟒𝟒𝟏𝟏,𝟏𝟏𝟏𝟏𝟒𝟒

𝐜𝐜𝐜𝐜 = 𝑩𝑩

𝟏𝟏𝟒𝟒𝟏𝟏𝟓𝟓


The depth of the water is 𝟏𝟏𝟒𝟒𝟏𝟏𝟓𝟓 𝐜𝐜𝐜𝐜.

b. How deep is the tank?

The depth of the water is 𝟏𝟏𝟑𝟑 the depth of the tank. Let 𝒅𝒅 represent the depth of the tank in centimeters.

𝟏𝟏𝟒𝟒𝟏𝟏𝟓𝟓

𝐜𝐜𝐜𝐜 =𝟏𝟏𝟑𝟑∙ 𝒅𝒅

𝟏𝟏𝟒𝟒𝟏𝟏𝟓𝟓

𝐜𝐜𝐜𝐜 ∙𝟑𝟑𝟏𝟏

=𝟏𝟏𝟑𝟑∙𝟑𝟑𝟏𝟏∙ 𝒅𝒅

𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜 +𝟑𝟑𝟓𝟓

𝐜𝐜𝐜𝐜 = 𝟏𝟏𝒅𝒅

𝟑𝟑𝟔𝟔𝟑𝟑𝟓𝟓

𝐜𝐜𝐜𝐜 = 𝒅𝒅

The depth of the tank is 𝟑𝟑𝟔𝟔𝟑𝟑𝟓𝟓 𝐜𝐜𝐜𝐜.

c. How many liters of water can the tank hold in total?



𝑽𝑽 = (𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏𝟖𝟖 𝐜𝐜𝐜𝐜) ∙ 𝟑𝟑𝟔𝟔𝟑𝟑𝟓𝟓

𝐜𝐜𝐜𝐜

𝑽𝑽 = 𝟏𝟏,𝟏𝟏𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝟑𝟑𝟔𝟔𝟑𝟑𝟓𝟓

𝐜𝐜𝐜𝐜

𝑽𝑽 = 𝟒𝟒𝟒𝟒,𝟑𝟑𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 + 𝟔𝟔𝟕𝟕𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑

𝑽𝑽 = 𝟒𝟒𝟒𝟒,𝟒𝟒𝟒𝟒𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑

𝟒𝟒𝟒𝟒,𝟒𝟒𝟒𝟒𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟒𝟒𝟒𝟒.𝟒𝟒𝟒𝟒𝟏𝟏 𝐋𝐋 The tank can hold up to 𝟒𝟒𝟏𝟏.𝟒𝟒 𝐋𝐋 of water.

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1. Mark wants to put some fish and decorative rocks in his new glass fish tank. He measured the outside dimensions of the right rectangular prism and recorded a length of 𝟓𝟓𝟓𝟓 𝐜𝐜𝐜𝐜, width of 𝟒𝟒𝟏𝟏 𝐜𝐜𝐜𝐜, and height of 𝟑𝟑𝟖𝟖 𝐜𝐜𝐜𝐜. He calculates that the tank will hold 𝟖𝟖𝟕𝟕.𝟕𝟕𝟖𝟖 𝐋𝐋 of water. Why is Mark’s calculation of volume incorrect? What is the correct volume? Mark also failed to take into account the fish and decorative rocks he plans to add. How will this affect the volume of water in the tank? Explain.

𝑽𝑽 = 𝑩𝑩𝑩𝑩 = (𝒍𝒍𝒍𝒍)𝑩𝑩

𝑽𝑽 = 𝟓𝟓𝟓𝟓 𝐜𝐜𝐜𝐜 ∙ 𝟒𝟒𝟏𝟏 𝐜𝐜𝐜𝐜 ∙ 𝟑𝟑𝟖𝟖 𝐜𝐜𝐜𝐜

𝑽𝑽 = 𝟏𝟏,𝟑𝟑𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝟑𝟑𝟖𝟖 𝐜𝐜𝐜𝐜

𝑽𝑽 = 𝟖𝟖𝟕𝟕,𝟕𝟕𝟖𝟖𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑

𝟖𝟖𝟕𝟕,𝟕𝟕𝟖𝟖𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟖𝟖𝟕𝟕.𝟕𝟕𝟖𝟖 𝐋𝐋

Mark measured only the outside dimensions of the fish tank and did not account for the thickness of the sides of the tank. If he fills the tank with 𝟖𝟖𝟕𝟕.𝟕𝟕𝟖𝟖 𝐋𝐋 of water, the water will overflow the sides. Mark also plans to put fish and rocks in the tank, which will force water out of the tank if it is filled to capacity.

2. Leondra bought an aquarium that is a right rectangular prism. The inside dimensions of the aquarium are 𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜 long, by 𝟒𝟒𝟖𝟖 𝐜𝐜𝐜𝐜 wide, by 𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜 deep. She plans to put water in the aquarium before purchasing any pet fish. How many liters of water does she need to put in the aquarium so that the water level is 𝟓𝟓 𝐜𝐜𝐜𝐜 below the top?

If the aquarium is 𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜 deep, then she wants the water to be 𝟓𝟓𝟓𝟓 𝐜𝐜𝐜𝐜 deep. Water takes on the shape of its container, so the water will form a right rectangular prism with a length of 𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜, a width of 𝟒𝟒𝟖𝟖 𝐜𝐜𝐜𝐜, and a height of 𝟓𝟓𝟓𝟓 𝐜𝐜𝐜𝐜.


𝑽𝑽 = (𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜 ∙ 𝟒𝟒𝟖𝟖 𝐜𝐜𝐜𝐜) ∙ 𝟓𝟓𝟓𝟓 𝐜𝐜𝐜𝐜

𝑽𝑽 = 𝟒𝟒,𝟑𝟑𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝟓𝟓𝟓𝟓 𝐜𝐜𝐜𝐜

𝑽𝑽 = 𝟏𝟏𝟑𝟑𝟕𝟕,𝟔𝟔𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑

𝟏𝟏𝟑𝟑𝟕𝟕,𝟔𝟔𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟏𝟏𝟑𝟑𝟕𝟕.𝟔𝟔 𝐋𝐋

The volume of water needed is 𝟏𝟏𝟑𝟑𝟕𝟕.𝟔𝟔 𝐋𝐋.

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3. The inside space of two different water tanks are shown below. Which tank has a greater capacity? Justify your answer.

𝑽𝑽𝟏𝟏 = 𝑩𝑩𝑩𝑩 = (𝒍𝒍𝒍𝒍)𝑩𝑩

𝑽𝑽𝟏𝟏 = �𝟔𝟔 𝐢𝐢𝐢𝐢. ∙ 𝟏𝟏𝟏𝟏𝟏𝟏

𝐢𝐢𝐢𝐢. � ∙ 𝟑𝟑 𝐢𝐢𝐢𝐢.

𝑽𝑽𝟏𝟏 = (𝟔𝟔 𝐢𝐢𝐢𝐢𝟏𝟏 + 𝟑𝟑 𝐢𝐢𝐢𝐢𝟏𝟏) ∙ 𝟑𝟑 𝐢𝐢𝐢𝐢.

𝑽𝑽𝟏𝟏 = 𝟒𝟒 𝐢𝐢𝐢𝐢𝟏𝟏 ∙ 𝟑𝟑 𝐢𝐢𝐢𝐢.

𝑽𝑽𝟏𝟏 = 𝟏𝟏𝟕𝟕 𝐢𝐢𝐢𝐢𝟑𝟑

𝑽𝑽𝟏𝟏 = 𝑩𝑩𝑩𝑩 = (𝒍𝒍𝒍𝒍)𝑩𝑩

𝑽𝑽𝟏𝟏 = �𝟏𝟏𝟏𝟏𝟏𝟏

𝐢𝐢𝐢𝐢. ∙ 𝟏𝟏 𝐢𝐢𝐢𝐢. � ∙ 𝟒𝟒 𝐢𝐢𝐢𝐢.

𝑽𝑽𝟏𝟏 = (𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏 + 𝟏𝟏 𝐢𝐢𝐢𝐢𝟏𝟏) ∙ 𝟒𝟒 𝐢𝐢𝐢𝐢 .

𝑽𝑽𝟏𝟏 = 𝟑𝟑 𝐢𝐢𝐢𝐢𝟏𝟏 ∙ 𝟒𝟒 𝐢𝐢𝐢𝐢.

𝑽𝑽𝟏𝟏 = 𝟏𝟏𝟕𝟕 𝐢𝐢𝐢𝐢𝟑𝟑

The tanks have the same volume, 𝟏𝟏𝟕𝟕 𝐢𝐢𝐢𝐢𝟑𝟑. Each prism has a face with an area of 𝟏𝟏𝟖𝟖 𝐢𝐢𝐢𝐢𝟏𝟏 (base) and a height that is

𝟏𝟏𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢.

4. The inside of a tank is in the shape of a right rectangular prism. The base of that prism is 𝟖𝟖𝟓𝟓 𝐜𝐜𝐜𝐜 by 𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜. What is the minimum height inside the tank if the volume of the liquid in the tank is 9𝟏𝟏 𝐋𝐋 ?


𝟒𝟒𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = (𝟖𝟖𝟓𝟓 𝐜𝐜𝐜𝐜 ∙ 𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜) ∙ 𝑩𝑩 𝟒𝟒𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟓𝟓,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟒𝟒𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 ∙𝟏𝟏

𝟓𝟓,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 = 𝟓𝟓,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙𝟏𝟏

𝟓𝟓,𝟒𝟒𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟒𝟒𝟏𝟏,𝟒𝟒𝟒𝟒𝟒𝟒𝟓𝟓,𝟒𝟒𝟒𝟒𝟒𝟒


𝟏𝟏𝟔𝟔𝟑𝟑𝟏𝟏𝟑𝟑𝟒𝟒


The minimum height of the inside of the tank is 𝟏𝟏𝟔𝟔𝟑𝟑𝟏𝟏𝟑𝟑𝟒𝟒 𝐜𝐜𝐜𝐜.

Tank 1

Tank 2

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7•3 Lesson 24


5. An oil tank is the shape of a right rectangular prism. The inside of the tank is 𝟑𝟑𝟔𝟔.𝟓𝟓 cm long, 𝟓𝟓𝟏𝟏 cm wide, and 𝟏𝟏𝟒𝟒 cm high. If 𝟒𝟒𝟓𝟓 liters of oil have been removed from the tank since it was full, what is the current depth of oil left in the tank?


𝑽𝑽 = (𝟑𝟑𝟔𝟔.𝟓𝟓 𝐜𝐜𝐜𝐜 ∙ 𝟓𝟓𝟏𝟏 𝐜𝐜𝐜𝐜) ∙ 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜

𝑽𝑽 = 𝟏𝟏,𝟖𝟖𝟒𝟒𝟖𝟖 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜

𝑽𝑽 = 𝟓𝟓𝟓𝟓,𝟒𝟒𝟒𝟒𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑

The tank has a capacity of 𝟓𝟓𝟓𝟓,𝟒𝟒𝟒𝟒𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑, or 𝟓𝟓𝟓𝟓.𝟒𝟒𝟒𝟒𝟏𝟏 𝐋𝐋.

𝟓𝟓𝟓𝟓.𝟒𝟒𝟒𝟒𝟏𝟏 𝐋𝐋− 𝟒𝟒𝟓𝟓 𝐋𝐋 = 𝟏𝟏𝟒𝟒.𝟒𝟒𝟒𝟒𝟏𝟏 𝐋𝐋

If 𝟒𝟒𝟓𝟓 𝐋𝐋 of oil have been removed from the tank, then 𝟏𝟏𝟒𝟒.𝟒𝟒𝟒𝟒𝟏𝟏 𝐋𝐋 are left in the tank.


𝟏𝟏𝟒𝟒,𝟒𝟒𝟒𝟒𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑 = (𝟑𝟑𝟔𝟔.𝟓𝟓 𝐜𝐜𝐜𝐜 ∙ 𝟓𝟓𝟏𝟏 𝐜𝐜𝐜𝐜) ∙ 𝑩𝑩 𝟏𝟏𝟒𝟒,𝟒𝟒𝟒𝟒𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟏𝟏,𝟖𝟖𝟒𝟒𝟖𝟖 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟏𝟏𝟒𝟒,𝟒𝟒𝟒𝟒𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑 ∙𝟏𝟏

𝟏𝟏,𝟖𝟖𝟒𝟒𝟖𝟖 𝐜𝐜𝐜𝐜𝟏𝟏 = 𝟏𝟏,𝟖𝟖𝟒𝟒𝟖𝟖 𝐜𝐜𝐜𝐜𝟏𝟏 ∙𝟏𝟏

𝟏𝟏,𝟖𝟖𝟒𝟒𝟖𝟖 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟏𝟏𝟒𝟒,𝟒𝟒𝟒𝟒𝟏𝟏𝟏𝟏,𝟖𝟖𝟒𝟒𝟖𝟖


𝟓𝟓.𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜 ≈ 𝑩𝑩

The depth of oil left in the tank is approximately 𝟓𝟓.𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜.

6. The inside of a right rectangular prism-shaped tank has a base that is 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜 by 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜 and a height of 𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜. The tank is filled to its capacity with water, and then 𝟏𝟏𝟒𝟒.𝟒𝟒𝟏𝟏 𝐋𝐋 of water is removed. How far did the water level drop?


𝑽𝑽 = (𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜) ∙ 𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜

𝑽𝑽 = 𝟑𝟑𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜

𝑽𝑽 = 𝟏𝟏𝟒𝟒,𝟏𝟏𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑

The capacity of the tank is 𝟏𝟏𝟒𝟒,𝟏𝟏𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 or 𝟏𝟏𝟒𝟒.𝟏𝟏𝟔𝟔 𝐋𝐋.

𝟏𝟏𝟒𝟒,𝟏𝟏𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 − 𝟏𝟏𝟒𝟒,𝟒𝟒𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟒𝟒,𝟏𝟏𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑

When 𝟏𝟏𝟒𝟒.𝟒𝟒𝟏𝟏 𝐋𝐋 or 𝟏𝟏𝟒𝟒,𝟒𝟒𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 of water is removed from the tank, there remains 𝟒𝟒,𝟏𝟏𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 of water in the tank.


𝟒𝟒,𝟏𝟏𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = (𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜 ∙ 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜) ∙ 𝑩𝑩 𝟒𝟒,𝟏𝟏𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟑𝟑𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟒𝟒,𝟏𝟏𝟒𝟒𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 ∙𝟏𝟏

𝟑𝟑𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏 = 𝟑𝟑𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏 ∙𝟏𝟏

𝟑𝟑𝟑𝟑𝟔𝟔 𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟒𝟒,𝟏𝟏𝟒𝟒𝟒𝟒𝟑𝟑𝟑𝟑𝟔𝟔


𝟏𝟏𝟕𝟕𝟏𝟏𝟏𝟏


The depth of the water left in the tank is 𝟏𝟏𝟕𝟕𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜.

𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜− 𝟏𝟏𝟕𝟕𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 = 𝟑𝟑𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜

This means that the water level has dropped 𝟑𝟑𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜.

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7•3 Lesson 24


7. A right rectangular prism-shaped container has inside dimensions of 𝟕𝟕𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 long and 𝟒𝟒𝟑𝟑𝟓𝟓 𝐜𝐜𝐜𝐜 wide. The tank is 𝟑𝟑𝟓𝟓 full of vegetable oil. It contains 𝟒𝟒.𝟒𝟒𝟏𝟏𝟒𝟒 𝐋𝐋 of oil. Find the height of the container.


𝟒𝟒𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = �𝟕𝟕𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜 ∙ 𝟒𝟒𝟑𝟑𝟓𝟓

𝐜𝐜𝐜𝐜� ∙ 𝑩𝑩

𝟒𝟒𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟑𝟑𝟒𝟒𝟏𝟏𝟏𝟏

𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟒𝟒𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 =𝟔𝟔𝟒𝟒𝟏𝟏

𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟒𝟒𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜𝟑𝟑 ∙𝟏𝟏

𝟔𝟔𝟒𝟒 𝐜𝐜𝐜𝐜𝟏𝟏 =𝟔𝟔𝟒𝟒𝟏𝟏

𝐜𝐜𝐜𝐜𝟏𝟏 ∙𝟏𝟏

𝟔𝟔𝟒𝟒𝐜𝐜𝐜𝐜𝟏𝟏 ∙ 𝑩𝑩

𝟖𝟖𝟏𝟏𝟖𝟖𝟔𝟔𝟒𝟒


𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 = 𝑩𝑩

The vegetable oil in the container is 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 deep, but this is only 𝟑𝟑𝟓𝟓 of the container’s depth. Let 𝒅𝒅 represent the depth of the container in centimeters.

𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 =𝟑𝟑𝟓𝟓∙ 𝒅𝒅

𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 ∙𝟓𝟓𝟑𝟑

=𝟑𝟑𝟓𝟓∙𝟓𝟓𝟑𝟑∙ 𝒅𝒅

𝟔𝟔𝟒𝟒𝟑𝟑

𝐜𝐜𝐜𝐜 = 𝟏𝟏 ∙ 𝒅𝒅

𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜 = 𝒅𝒅

The depth of the container is 𝟏𝟏𝟒𝟒 𝐜𝐜𝐜𝐜.

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7•3 Lesson 24


8. A right rectangular prism with length of 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢, width of 𝟏𝟏𝟔𝟔 𝐢𝐢𝐢𝐢, and height of 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 is 𝟏𝟏𝟑𝟑 filled with water. If the water is emptied into another right rectangular prism with a length of 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢, a width of 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢, and height of 𝟒𝟒 𝐢𝐢𝐢𝐢, will the second container hold all of the water? Explain why or why not. Determine how far (above or below) the water level would be from the top of the container.

𝟏𝟏𝟑𝟑∙ 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 =

𝟏𝟏𝟒𝟒𝟑𝟑

𝐢𝐢𝐢𝐢 = 𝟖𝟖 𝐢𝐢𝐢𝐢 The height of the water in the first prism is 𝟖𝟖 𝐢𝐢𝐢𝐢.


𝑽𝑽 = (𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢 ∙ 𝟏𝟏𝟔𝟔 𝐢𝐢𝐢𝐢) ∙ 𝟖𝟖 𝐢𝐢𝐢𝐢

𝑽𝑽 = 𝟏𝟏𝟔𝟔𝟒𝟒 𝐢𝐢𝐢𝐢𝟏𝟏 ∙ 𝟖𝟖 𝐢𝐢𝐢𝐢

𝑽𝑽 = 𝟏𝟏,𝟏𝟏𝟖𝟖𝟒𝟒 𝐢𝐢𝐢𝐢𝟑𝟑

The volume of water is 𝟏𝟏,𝟏𝟏𝟖𝟖𝟒𝟒 𝐢𝐢𝐢𝐢𝟑𝟑.


𝑽𝑽 = (𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 ∙ 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢) ∙ 𝟒𝟒 𝐢𝐢𝐢𝐢

𝑽𝑽 = 𝟏𝟏𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢𝟏𝟏 ∙ 𝟒𝟒 𝐢𝐢𝐢𝐢

𝑽𝑽 = 𝟏𝟏,𝟏𝟏𝟒𝟒𝟔𝟔 𝐢𝐢𝐢𝐢𝟑𝟑

The capacity of the second prism is 𝟏𝟏,𝟏𝟏𝟒𝟒𝟔𝟔 𝐢𝐢𝐢𝐢𝟑𝟑, which is greater than the volume of water, so the water will fit in the second prism.

𝑽𝑽 = 𝑩𝑩𝑩𝑩 = (𝒍𝒍𝒍𝒍)𝑩𝑩 Let 𝑩𝑩 represent the depth of the water in the second prism in inches.

𝟏𝟏,𝟏𝟏𝟖𝟖𝟒𝟒 𝐢𝐢𝐢𝐢𝟑𝟑 = (𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 ∙ 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢) ∙ 𝑩𝑩

𝟏𝟏,𝟏𝟏𝟖𝟖𝟒𝟒 𝐢𝐢𝐢𝐢𝟑𝟑 = (𝟏𝟏𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢𝟏𝟏) ∙ 𝑩𝑩

𝟏𝟏,𝟏𝟏𝟖𝟖𝟒𝟒 𝐢𝐢𝐢𝐢𝟑𝟑 ∙𝟏𝟏

𝟏𝟏𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢𝟏𝟏= 𝟏𝟏𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢𝟏𝟏 ∙

𝟏𝟏𝟏𝟏𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢𝟏𝟏

∙ 𝑩𝑩

𝟏𝟏,𝟏𝟏𝟖𝟖𝟒𝟒𝟏𝟏𝟒𝟒𝟒𝟒

𝐢𝐢𝐢𝐢 = 𝟏𝟏 ∙ 𝑩𝑩

𝟖𝟖𝟏𝟏𝟏𝟏𝟖𝟖𝟏𝟏𝟒𝟒𝟒𝟒

𝐢𝐢𝐢𝐢 = 𝑩𝑩

𝟖𝟖𝟖𝟖𝟒𝟒 𝐢𝐢𝐢𝐢 = 𝑩𝑩

The depth of the water in the second prism is 𝟖𝟖𝟖𝟖𝟒𝟒 𝐢𝐢𝐢𝐢.

𝟒𝟒 𝐢𝐢𝐢𝐢 − 𝟖𝟖𝟖𝟖𝟒𝟒 𝐢𝐢𝐢𝐢 = 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢

The water level will be 𝟏𝟏𝟒𝟒

𝐢𝐢𝐢𝐢 from the top of the second prism.

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7•3 Lesson 25

Lesson 25: Volume and Surface Area


Student Outcomes

Students solve real-world and mathematical problems involving volume and surface areas of three-dimensional objects composed of cubes and right prisms.

Lesson Notes In this lesson, students apply what they learned in Lessons 22–24 to solve real-world problems. As students work on the problems, encourage them to present their approaches for determining the volume and surface area. The initial questions specifically ask for volume, but later in the lesson, students must interpret the context of the problem to know which measurement to choose. Several problems involve finding the height of a prism if the volume and two other dimensions are given. Students work with cubic units and units of liquid measure on the volume problems.

Classwork

Opening (2 minutes)

In the Opening Exercise, students are asked to find the volume and surface area of a right rectangular prism. This exercise provides information about students who may need some additional support during the lesson if they have difficulty solving this problem. Tell the class that today they are applying what they learned about finding the surface area and volume of prisms to real-world problems.


What is the surface area and volume of the right rectangular prism?

Surface Area = 𝟐𝟐(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢)(𝟔𝟔.𝟓𝟓 𝐢𝐢𝐢𝐢) + 𝟐𝟐(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢)(𝟔𝟔.𝟓𝟓 𝐢𝐢𝐢𝐢) + 𝟐𝟐(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢)(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢) = 𝟒𝟒𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢𝟐𝟐

Volume = 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 ⋅ 𝟔𝟔.𝟓𝟓 𝐢𝐢𝐢𝐢 ∙ 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 = 𝟕𝟕𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒

𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢

𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 𝟔𝟔.𝟓𝟓 𝐢𝐢𝐢𝐢

Scaffolding: This lesson builds gradually to more and more complicated problems. Provide additional practice at each stage if you find students are struggling.

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Example 1 (10 minutes): Volume of a Fish Tank

This example uses the prism from the Opening Exercise and applies it to a real-world situation. Either guide students through this example, allow them to work with a partner, or allow them to work in small groups, depending on their level. If you have students work with a partner or a group, be sure to present different solutions and monitor the groups’ progress.

For part (a) below.

How did you identify this as a volume problem?

The term gallon refers to capacity or volume.

Be sure that students recognize the varying criteria for calculating surface area and volume.

For part (c) below.

What helped you to understand that this is a surface area problem? Square inches are measures of area, not volume.

Covering the sides requires using an area calculation, not a volume calculation.

Example 1: Volume of a Fish Tank

Jay has a small fish tank. It is the same shape and size as the right rectangular prism shown in the Opening Exercise.

a. The box it came in says that it is a 𝟒𝟒-gallon tank. Is this claim true? Explain your reasoning. Recall that 𝟏𝟏 𝐠𝐠𝐠𝐠𝐠𝐠 = 𝟐𝟐𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢𝟒𝟒.

The volume of the tank is 𝟕𝟕𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒. To convert cubic inches to gallons, divide by 𝟐𝟐𝟒𝟒𝟏𝟏.

𝟕𝟕𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒 ∙𝟏𝟏 𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐢𝐢𝟐𝟐𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢𝟒𝟒

= 𝟒𝟒.𝟏𝟏𝟒𝟒 𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐢𝐢𝐠𝐠

The claim is true if you round to the nearest whole gallon.

b. The pet store recommends filling the tank to within 𝟏𝟏.𝟓𝟓 𝐢𝐢𝐢𝐢 of the top. How many gallons of water will the tank hold if it is filled to the recommended level?

Use 𝟖𝟖.𝟓𝟓 𝐢𝐢𝐢𝐢. instead of 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. to calculate the volume. 𝑽𝑽 = 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 ∙ 𝟔𝟔.𝟓𝟓 𝐢𝐢𝐢𝐢 ∙ 𝟖𝟖.𝟓𝟓 𝐢𝐢𝐢𝐢 = 𝟔𝟔𝟏𝟏𝟕𝟕.𝟕𝟕𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒.

𝟔𝟔𝟏𝟏𝟕𝟕.𝟕𝟕𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒 ∙𝟏𝟏 𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐢𝐢𝟐𝟐𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢𝟒𝟒

= 𝟐𝟐.𝟔𝟔𝟒𝟒 𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐢𝐢𝐠𝐠

c. Jay wants to cover the back, left, and right sides of the tank with a background picture. How many square inches will be covered by the picture?

Back side area = 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 ∙ 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐

Left and right side area = 𝟐𝟐(𝟔𝟔.𝟓𝟓 𝐢𝐢𝐢𝐢)(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢) = 𝟏𝟏𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐

The total area to be covered with the background picture is 𝟐𝟐𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐.

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7•3 Lesson 25


d. Water in the tank evaporates each day, causing the water level to drop. How many gallons of water have evaporated by the time the water in the tank is four inches deep? Assume the tank was filled to within 𝟏𝟏.𝟓𝟓 𝐢𝐢𝐢𝐢. of the top to start.

Volume when water is 𝟒𝟒 𝐢𝐢𝐢𝐢. deep:

𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢.∙ 𝟔𝟔.𝟓𝟓 𝐢𝐢𝐢𝐢.∙ 𝟒𝟒 𝐢𝐢𝐢𝐢 = 𝟐𝟐𝟖𝟖𝟔𝟔 𝐢𝐢𝐢𝐢𝟒𝟒

Difference in the two volumes:

𝟔𝟔𝟏𝟏𝟕𝟕.𝟕𝟕𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒 − 𝟐𝟐𝟖𝟖𝟔𝟔 𝐢𝐢𝐢𝐢𝟒𝟒 = 𝟒𝟒𝟐𝟐𝟏𝟏.𝟕𝟕𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒

When the water is filled to within 𝟏𝟏.𝟓𝟓 𝐢𝐢𝐢𝐢 of the top, the volume is 𝟔𝟔𝟏𝟏𝟕𝟕.𝟕𝟕𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒. When the water is 𝟒𝟒 𝐢𝐢𝐢𝐢 deep, the volume is 𝟐𝟐𝟖𝟖𝟔𝟔 𝐢𝐢𝐢𝐢𝟒𝟒. The difference in the two volumes is 𝟒𝟒𝟐𝟐𝟏𝟏.𝟕𝟕𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒. Converting cubic inches to gallons by dividing by 𝟐𝟐𝟒𝟒𝟏𝟏 gives a difference of 𝟏𝟏.𝟒𝟒𝟒𝟒 𝐠𝐠𝐠𝐠𝐠𝐠., which means 𝟏𝟏.𝟒𝟒𝟒𝟒 𝐠𝐠𝐠𝐠𝐠𝐠. of water have evaporated.

Use these questions with the whole class or small groups as discussion points.

Which problems involve measuring the surface area? Which problems involve measuring the volume?

Covering the sides of the tank involved surface area. The other problems asked about the amount of water the tank would hold, which required us to measure the volume of the tank.

How do you convert cubic inches to gallons?

You need to divide the total cubic inches by the number of cubic inches in one gallon. What are some different ways to answer part (c)?

Answers will vary. You could do each side separately, or you could do the left side and multiply it by 2, and then add the area of the back side.

Exercise 1 (10 minutes): Fish Tank Designs

In this exercise, students compare the volume of two different right prisms. They consider the differences in the surface areas and volumes of differently shaped tanks. This example presents two solid figures where a figure with larger volume has a smaller surface area. In part (c), students explore whether or not this is always true. After completing the exercise, have the class consider the following questions as you discuss this exercise. If time permits, encourage students to consider how a company that manufactures fish tanks might decide on its designs. Encourage students to make claims and respond to the claims of others. Below are some possible discussion questions to pose to students after the exercises are completed.

When comparing the volumes and the surface areas, the larger-volume tank has the smaller surface area. Why? Will it always be like that?

Changing the dimensions of the base affects the surface area. Shapes that are more like a cube will have a smaller surface area. For a rectangular base tank, where the area of the base is a long and skinny rectangle, the surface area is much greater. For example, a tank with a base that is 50 in. by 5 in. has a surface area of 2(5 in)(50 in) + 2(5 in)(15 in) + 2(50 in)(15 in), or 2150 in2. The surface area is more than the trapezoid base tank, but the volume is the same.

Why might a company be interested in building a fish tank that has a smaller surface area for a larger volume? What other parts of the design might make a difference when building a fish tank? The company that makes tanks might set its prices based on the amount of material used. If the

volumes are the same, then the tank with fewer materials would be cheaper to make. The company might make designs that are more interesting to buyers, such as the trapezoidal prism.

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7•3 Lesson 25


Exercise 1: Fish Tank Designs

Two fish tanks are shown below, one in the shape of a right rectangular prism (𝑹𝑹) and one in the shape of a right trapezoidal prism (𝑻𝑻).

a. Which tank holds the most water? Let 𝑽𝑽𝑽𝑽𝑽𝑽(𝑹𝑹) represent the volume of the right rectangular prism and 𝑽𝑽𝑽𝑽𝑽𝑽(𝑻𝑻) represent the volume of the right trapezoidal prism. Use your answer to fill in the blanks with 𝑽𝑽𝑽𝑽𝑽𝑽(𝑹𝑹) and 𝑽𝑽𝑽𝑽𝑽𝑽(𝑻𝑻).

Volume of the right rectangular prism: (𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢 × 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢) × 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢 = 𝟒𝟒,𝟕𝟕𝟓𝟓𝟏𝟏 𝐢𝐢𝐢𝐢𝟒𝟒

Volume of the right trapezoidal prism: (𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢× 𝟖𝟖 𝐢𝐢𝐢𝐢) × 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢 = 𝟒𝟒,𝟕𝟕𝟐𝟐𝟏𝟏 𝐢𝐢𝐢𝐢𝟒𝟒

The right rectangular prism holds the most water.

𝑽𝑽𝑽𝑽𝑽𝑽(𝑻𝑻) < 𝑽𝑽𝑽𝑽𝑽𝑽(𝑹𝑹)

b. Which tank has the most surface area? Let 𝑺𝑺𝑺𝑺(𝑹𝑹) represent the surface area of the right rectangular prism and 𝑺𝑺𝑺𝑺(𝑻𝑻) represent the surface area of the right trapezoidal prism. Use your answer to fill in the blanks with 𝑺𝑺𝑺𝑺(𝑹𝑹) and 𝑺𝑺𝑺𝑺(𝑻𝑻).

The surface area of the right rectangular prism: 𝟐𝟐(𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢× 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢) + 𝟐𝟐(𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢× 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢) + 𝟐𝟐(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢× 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢) = 𝟓𝟓𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟕𝟕𝟓𝟓𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟒𝟒𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐 = 𝟏𝟏,𝟓𝟓𝟓𝟓𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐

The surface area of the right trapezoidal prism: 𝟐𝟐(𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢× 𝟖𝟖 𝐢𝐢𝐢𝐢) + 𝟐𝟐(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢× 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢) + (𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢× 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢) + (𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢× 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢)

= 𝟒𝟒𝟒𝟒𝟔𝟔 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟒𝟒𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟒𝟒𝟕𝟕𝟓𝟓 𝐢𝐢𝐢𝐢𝟐𝟐 + 𝟓𝟓𝟓𝟓𝟓𝟓 𝐢𝐢𝐢𝐢𝟐𝟐 = 𝟏𝟏𝟕𝟕𝟐𝟐𝟔𝟔 𝐢𝐢𝐢𝐢𝟐𝟐

The right trapezoidal prism has the most surface area.

𝑺𝑺𝑺𝑺(𝑹𝑹) < 𝑺𝑺𝑺𝑺(𝑻𝑻)

c. Water evaporates from each aquarium. After the water level has dropped 𝟏𝟏𝟐𝟐 inch in each aquarium, how many cubic inches of water are required to fill up each aquarium? Show work to support your answers.

The right rectangular prism will need 𝟏𝟏𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒 of water. The right trapezoidal prism will need 𝟏𝟏𝟐𝟐𝟒𝟒 𝐢𝐢𝐢𝐢𝟒𝟒 of water. First, decrease the height of each prism by a half inch and recalculate the volumes. Then, subtract each answer from the original volume of each prism.

𝐍𝐍𝐍𝐍𝐍𝐍𝐍𝐍𝐠𝐠𝑽𝑽(𝑹𝑹) = (𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢)(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢)(𝟏𝟏𝟒𝟒.𝟓𝟓 𝐢𝐢𝐢𝐢) = 𝟒𝟒,𝟔𝟔𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒

𝐍𝐍𝐍𝐍𝐍𝐍𝐍𝐍𝐠𝐠𝐠𝐠(𝑻𝑻) = (𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢)(𝟖𝟖 𝐢𝐢𝐢𝐢)(𝟏𝟏𝟒𝟒.𝟓𝟓 𝐢𝐢𝐢𝐢) = 𝟒𝟒,𝟓𝟓𝟒𝟒𝟔𝟔 𝐢𝐢𝐢𝐢𝟒𝟒

𝟒𝟒,𝟕𝟕𝟓𝟓𝟏𝟏 𝐢𝐢𝐢𝐢𝟒𝟒 − 𝟒𝟒,𝟔𝟔𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒 = 𝟏𝟏𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒 𝟒𝟒,𝟕𝟕𝟐𝟐𝟏𝟏 𝐢𝐢𝐢𝐢𝟒𝟒 − 𝟒𝟒,𝟓𝟓𝟒𝟒𝟔𝟔 𝐢𝐢𝐢𝐢𝟒𝟒 = 𝟏𝟏𝟐𝟐𝟒𝟒 𝐢𝐢𝐢𝐢𝟒𝟒

𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢.

𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. 𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢.

𝟔𝟔 𝐢𝐢𝐢𝐢.

𝟖𝟖 𝐢𝐢𝐢𝐢.

𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢.

𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢.

𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢. 𝟖𝟖 𝐢𝐢𝐢𝐢.

Tank 𝐑𝐑 Tank 𝐓𝐓

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Exercise 2 (15 minutes): Design Your Own Fish Tank

This is a very open-ended task. If students have struggled with the first example and exercise, you may wish to move them directly to some of the Problem Set exercises. Three possible solutions are presented below, but there are others. None of these solutions has a volume of exactly 10 gallons. Encourage students to find reasonable dimensions that are close to 10 gallons. The volume in cubic inches of a 10 gallon tank is 2310 in3. Students may try various approaches to this problem. Encourage them to select values for the dimensions of the tank that are realistic. For example, a rectangular prism tank that is 23 in by 20 in by 5 in is probably not a reasonable choice, even though the volume is exactly 2310 in3.

Exercise 2: Design Your Own Fish Tank

Design at least three fish tanks that will hold approximately 𝟏𝟏𝟏𝟏 gallons of water. All of the tanks should be shaped like right prisms. Make at least one tank have a base that is not a rectangle. For each tank, make a sketch, and calculate the volume in gallons to the nearest hundredth.

Three possible designs are shown below.

𝟏𝟏𝟏𝟏 𝐠𝐠𝐠𝐠𝐠𝐠. is 𝟐𝟐,𝟒𝟒𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢𝟒𝟒

Rectangular Base: Volume = 𝟐𝟐,𝟒𝟒𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢𝟒𝟒 or 𝟒𝟒.𝟒𝟒𝟕𝟕 𝐠𝐠𝐠𝐠𝐠𝐠.

Triangular Base: Volume = 𝟐𝟐,𝟐𝟐𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢𝟒𝟒 or 𝟒𝟒.𝟕𝟕𝟏𝟏 𝐠𝐠𝐠𝐠𝐠𝐠.

Hexagonal Base: Volume = 𝟐𝟐,𝟒𝟒𝟐𝟐𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒 or 𝟏𝟏𝟏𝟏.𝟏𝟏𝟔𝟔 𝐠𝐠𝐠𝐠𝐠𝐠.

Challenge: Each tank is to be constructed from glass that is 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢. thick. Select one tank that you designed, and determine the difference between the volume of the total tank (including the glass) and the volume inside the tank. Do not include a glass top on your tank.

Height = 𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢 − 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢 = 𝟏𝟏𝟏𝟏.𝟕𝟕𝟓𝟓 𝐢𝐢𝐢𝐢

Length = 𝟐𝟐𝟒𝟒 𝐢𝐢𝐢𝐢 − 𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢 = 𝟐𝟐𝟒𝟒.𝟓𝟓 𝐢𝐢𝐢𝐢

Width = 𝟖𝟖 𝐢𝐢𝐢𝐢 − 𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢 = 𝟕𝟕.𝟓𝟓 𝐢𝐢𝐢𝐢

Inside Volume = 𝟐𝟐,𝟏𝟏𝟕𝟕𝟏𝟏.𝟒𝟒 𝐢𝐢𝐢𝐢𝟒𝟒

The difference between the two volumes is 𝟐𝟐𝟒𝟒𝟒𝟒.𝟏𝟏 𝐢𝐢𝐢𝐢𝟒𝟒, which is approximately 𝟏𝟏 𝐠𝐠𝐠𝐠𝐠𝐠.

MP.4

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7•3 Lesson 25


Closing (2 minutes)

When discussing the third bulleted item below with students, emphasize the point by using two containers with the same volume (perhaps a rectangular cake pan and a more cube-like container). Pour water (or rice) into both. Ask students which container has a greater surface area (the rectangular cake pan). Then, pour the contents of the two containers into separate one-gallon milk jugs to see that, while the surface areas are different, the volume held by each is the same.

When the water is removed from a right prism-shaped tank, and the volume of water is reduced, which other measurement(s) also change? Which measurement(s) stay the same?

The height is also reduced, but the area of the base stays the same. How do you decide whether a problem asks you to find the surface area or the volume of a solid figure?

The decision is based on whether you are measuring the area of the sides of the solid or whether you are measuring the space inside. If you are filling a tank with water or a liquid, then the question is about volume. If you are talking about the materials required to build the solid, then the question is about surface area.

Does a bigger volume always mean a bigger surface area?

No. Two right prisms can have the same volume but different surface areas. If you increase the volume by making the shape more like a cube, the surface area might be less than if it were as a solid with a smaller volume.


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7•3 Lesson 25


15 in

30 in 48 in

Name Date


Exit Ticket Melody is planning a raised bed for her vegetable garden.

a. How many square feet of wood does she need to create the bed?

b. She needs to add soil. Each bag contains 1.5 cubic feet. How many bags will she need to fill the vegetable

garden?

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7•3 Lesson 25


𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢

𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢

𝟒𝟒𝟖𝟖 𝐢𝐢𝐢𝐢


Melody is planning a raised bed for her vegetable garden.

a. How many square feet of wood does she need to create the bed?

𝟐𝟐(𝟒𝟒 𝐟𝐟𝐟𝐟)(𝟏𝟏.𝟐𝟐𝟓𝟓 𝐟𝐟𝐟𝐟) + 𝟐𝟐(𝟐𝟐.𝟓𝟓 𝐟𝐟𝐟𝐟)(𝟏𝟏.𝟐𝟐𝟓𝟓 𝐟𝐟𝐟𝐟) = 𝟏𝟏𝟔𝟔.𝟐𝟐𝟓𝟓 𝐟𝐟𝐟𝐟𝟐𝟐

The dimensions in feet are 𝟒𝟒 𝐟𝐟𝐟𝐟. by 𝟏𝟏.𝟐𝟐𝟓𝟓 𝐟𝐟𝐟𝐟. by 𝟐𝟐.𝟓𝟓 𝐟𝐟𝐟𝐟. The lateral area is 𝟏𝟏𝟔𝟔.𝟐𝟐𝟓𝟓 𝐟𝐟𝐟𝐟𝟐𝟐.

b. She needs to add soil. Each bag contains 𝟏𝟏.𝟓𝟓 cubic feet. How many bags will she need to fill the vegetable garden?

𝑽𝑽 = 𝟒𝟒 𝐟𝐟𝐟𝐟 ∙ 𝟏𝟏.𝟐𝟐𝟓𝟓 𝐟𝐟𝐟𝐟 ∙ 𝟐𝟐.𝟓𝟓 𝐟𝐟𝐟𝐟 = 𝟏𝟏𝟐𝟐.𝟓𝟓 𝐟𝐟𝐟𝐟𝟒𝟒

The volume is 𝟏𝟏𝟐𝟐.𝟓𝟓 𝐟𝐟𝐟𝐟𝟒𝟒. Divide the total cubic feet by 𝟏𝟏.𝟓𝟓 𝐟𝐟𝐟𝐟𝟒𝟒 to determine the number of bags.

𝟏𝟏𝟐𝟐.𝟓𝟓 𝐟𝐟𝐟𝐟𝟒𝟒 ÷ 𝟏𝟏.𝟓𝟓 𝐟𝐟𝐟𝐟𝟒𝟒 = 𝟖𝟖𝟏𝟏𝟒𝟒

Melody will need to purchase 𝟒𝟒 bags of soil to fill the garden bed.

Note that if students fail to recognize the need to round up to nine bags, this should be addressed. Also, if the thickness of the wood were given, then there would be soil left over, and possibly only 8 bags would be needed, depending on the thickness.


1. The dimensions of several right rectangular fish tanks are listed below. Find the volume in cubic centimeters, the capacity in liters (𝟏𝟏 𝐋𝐋 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒), and the surface area in square centimeters for each tank. What do you observe about the change in volume compared with the change in surface area between the small tank and the extra-large tank?

Tank Size Length (𝐜𝐜𝐦𝐦) Width (𝐜𝐜𝐦𝐦) Height (𝐜𝐜𝐦𝐦) Small 𝟐𝟐𝟒𝟒 𝟏𝟏𝟖𝟖 𝟏𝟏𝟓𝟓

Medium 𝟒𝟒𝟏𝟏 𝟐𝟐𝟏𝟏 𝟐𝟐𝟏𝟏 Large 𝟒𝟒𝟔𝟔 𝟐𝟐𝟒𝟒 𝟐𝟐𝟓𝟓

Extra-Large 𝟒𝟒𝟏𝟏 𝟐𝟐𝟕𝟕 𝟒𝟒𝟏𝟏

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7•3 Lesson 25


𝟏𝟏𝟓𝟓 𝐜𝐜𝐦𝐦

𝟐𝟐𝟓𝟓 𝐜𝐜𝐦𝐦

Tank Size Volume (𝐜𝐜𝐦𝐦𝟒𝟒) Capacity (𝐋𝐋) Surface Area (𝐜𝐜𝐦𝐦𝟐𝟐) Small 𝟔𝟔,𝟒𝟒𝟖𝟖𝟏𝟏 𝟔𝟔.𝟒𝟒𝟖𝟖 𝟐𝟐,𝟏𝟏𝟐𝟐𝟒𝟒

Medium 𝟏𝟏𝟐𝟐,𝟔𝟔𝟏𝟏𝟏𝟏 𝟏𝟏𝟐𝟐.𝟔𝟔 𝟒𝟒,𝟒𝟒𝟏𝟏𝟏𝟏 Large 𝟐𝟐𝟏𝟏,𝟔𝟔𝟏𝟏𝟏𝟏 𝟐𝟐𝟏𝟏.𝟔𝟔 𝟒𝟒,𝟕𝟕𝟐𝟐𝟖𝟖

Extra-Large 𝟒𝟒𝟐𝟐,𝟒𝟒𝟏𝟏𝟏𝟏 𝟒𝟒𝟐𝟐.𝟒𝟒 𝟔𝟔,𝟏𝟏𝟖𝟖𝟏𝟏

While the volume of the extra-large tank is about five times the volume of the small tank, its surface area is less than three times that of the small tank.

2. A rectangular container 𝟏𝟏𝟓𝟓 𝐜𝐜𝐦𝐦 long by 𝟐𝟐𝟓𝟓 𝐜𝐜𝐦𝐦 wide contains 𝟐𝟐.𝟓𝟓 𝐋𝐋 of water.

a. Find the height of the water level in the container. (𝟏𝟏 𝐋𝐋 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒)

𝟐𝟐.𝟓𝟓 𝐋𝐋 = 𝟐𝟐,𝟓𝟓𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒

To find the height of the water level, divide the volume in cubic centimeters by the area of the base.

𝟐𝟐,𝟓𝟓𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒

𝟐𝟐𝟓𝟓 𝐜𝐜𝐦𝐦 ∙ 𝟏𝟏𝟓𝟓 𝐜𝐜𝐦𝐦= 𝟔𝟔

𝟐𝟐𝟒𝟒𝐜𝐜𝐦𝐦

b. If the height of the container is 𝟏𝟏𝟖𝟖 𝐜𝐜𝐦𝐦, how many more liters of water would it take to completely fill the container?

Volume of tank: (𝟐𝟐𝟓𝟓 𝐜𝐜𝐦𝐦× 𝟏𝟏𝟓𝟓 𝐜𝐜𝐦𝐦) × 𝟏𝟏𝟖𝟖 𝐜𝐜𝐦𝐦 = 𝟔𝟔,𝟕𝟕𝟓𝟓𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒

Capacity of tank: 𝟔𝟔.𝟕𝟕𝟓𝟓 𝐋𝐋

Difference: 𝟔𝟔.𝟕𝟕𝟓𝟓 𝐋𝐋 − 𝟐𝟐.𝟓𝟓 𝐋𝐋 = 𝟒𝟒.𝟐𝟐𝟓𝟓 𝐋𝐋

c. What percentage of the tank is filled when it contains 𝟐𝟐.𝟓𝟓 𝐋𝐋 of water?

𝟐𝟐.𝟓𝟓 𝐋𝐋𝟔𝟔.𝟕𝟕𝟓𝟓 𝐋𝐋

= 𝟏𝟏.𝟒𝟒𝟕𝟕 = 𝟒𝟒𝟕𝟕%

3. A rectangular container measuring 𝟐𝟐𝟏𝟏 𝐜𝐜𝐦𝐦 by 𝟏𝟏𝟒𝟒.𝟓𝟓 𝐜𝐜𝐦𝐦 by 𝟏𝟏𝟏𝟏.𝟓𝟓 𝐜𝐜𝐦𝐦 is filled with water to its brim. If 𝟒𝟒𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒 are drained out of the container, what will be the height of the water level? If necessary, round to the nearest tenth.

Volume: (𝟐𝟐𝟏𝟏 𝐜𝐜𝐦𝐦× 𝟏𝟏𝟒𝟒.𝟓𝟓 𝐜𝐜𝐦𝐦) × 𝟏𝟏𝟏𝟏.𝟓𝟓 𝐜𝐜𝐦𝐦 = 𝟒𝟒,𝟏𝟏𝟒𝟒𝟓𝟓 𝐜𝐜𝐦𝐦𝟒𝟒

Volume after draining: 𝟐𝟐,𝟕𝟕𝟒𝟒𝟓𝟓 𝐜𝐜𝐦𝐦𝟒𝟒

Height (divide the volume by the area of the base):

𝟐𝟐𝟕𝟕𝟒𝟒𝟓𝟓 𝐜𝐜𝐦𝐦𝟒𝟒

𝟐𝟐𝟏𝟏 𝐜𝐜𝐦𝐦 × 𝟏𝟏𝟒𝟒.𝟓𝟓 𝐜𝐜𝐦𝐦≈ 𝟒𝟒.𝟓𝟓 𝐜𝐜𝐦𝐦

𝟏𝟏𝟏𝟏.𝟓𝟓 𝐜𝐜𝐦𝐦

? 𝟏𝟏𝟒𝟒.𝟓𝟓 𝐜𝐜𝐦𝐦

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7•3Lesson 25


𝟕𝟕𝟓𝟓 𝐜𝐜𝐦𝐦

𝟖𝟖𝟓𝟓 𝐜𝐜𝐦𝐦

𝟒𝟒𝟏𝟏 𝐜𝐜𝐦𝐦

𝟔𝟔𝟏𝟏 𝐜𝐜𝐦𝐦

𝟔𝟔𝟏𝟏 𝐜𝐜𝐦𝐦 𝟒𝟒𝟏𝟏 𝐜𝐜𝐦𝐦

𝟓𝟓𝟓𝟓 𝐜𝐜𝐦𝐦

𝟓𝟓𝟏𝟏 𝐜𝐜𝐦𝐦

𝟒𝟒𝟓𝟓 𝐜𝐜𝐦𝐦

𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦

𝟐𝟐𝟕𝟕 𝐜𝐜𝐦𝐦

𝟒𝟒𝟓𝟓 𝐜𝐜𝐦𝐦

4. Two tanks are shown below. Both are filled to capacity, but the owner decides to drain them. Tank 1 is draining at a rate of 𝟖𝟖 liters per minute. Tank 2 is draining at a rate of 𝟏𝟏𝟏𝟏 liters per minute. Which tank empties first?

Tank 1 Tank 2

Tank 1 Volume: 𝟕𝟕𝟓𝟓 𝐜𝐜𝐦𝐦 × 𝟔𝟔𝟏𝟏 𝐜𝐜𝐦𝐦 × 𝟔𝟔𝟏𝟏 𝐜𝐜𝐦𝐦 = 𝟐𝟐𝟕𝟕𝟏𝟏,𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒

Tank 2 Volume: 𝟒𝟒𝟏𝟏 𝐜𝐜𝐦𝐦 × 𝟒𝟒𝟏𝟏 𝐜𝐜𝐦𝐦 × 𝟖𝟖𝟓𝟓 𝐜𝐜𝐦𝐦 = 𝟒𝟒𝟏𝟏𝟔𝟔,𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒

Tank 1 Capacity: 𝟐𝟐𝟕𝟕𝟏𝟏 𝐋𝐋 Tank 2 Capacity: 𝟒𝟒𝟏𝟏𝟔𝟔 𝐋𝐋

To find the time to drain each tank, divide the capacity by the rate (liters per minute).

Time to drain tank 1: 𝟐𝟐𝟕𝟕𝟏𝟏 𝐋𝐋

𝟖𝟖 𝐋𝐋𝐦𝐦𝐢𝐢𝐢𝐢

= 𝟒𝟒𝟒𝟒. 𝟕𝟕𝟓𝟓 𝐦𝐦𝐢𝐢𝐢𝐢. Time to drain tank 2: 𝟒𝟒𝟏𝟏𝟔𝟔 𝐋𝐋

𝟏𝟏𝟏𝟏 𝐋𝐋𝐦𝐦𝐢𝐢𝐢𝐢

= 𝟒𝟒𝟏𝟏. 𝟔𝟔 𝐦𝐦𝐢𝐢𝐢𝐢.

Tank 2 empties first.

5. Two tanks are shown below. One tank is draining at a rate of 𝟖𝟖 liters per minute into the other one, which is empty. After 𝟏𝟏𝟏𝟏 minutes, what will be the height of the water level in the second tank? If necessary, round to the nearest minute.

Volume of the top tank: 𝟒𝟒𝟓𝟓 𝐜𝐜𝐦𝐦 × 𝟓𝟓𝟏𝟏 𝐜𝐜𝐦𝐦× 𝟓𝟓𝟓𝟓 𝐜𝐜𝐦𝐦 = 𝟏𝟏𝟐𝟐𝟒𝟒,𝟕𝟕𝟓𝟓𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒

Capacity of the top tank: 𝟏𝟏𝟐𝟐𝟒𝟒.𝟕𝟕𝟓𝟓 𝐋𝐋

At 𝟖𝟖 𝐋𝐋𝐦𝐦𝐢𝐢𝐢𝐢 for 𝟏𝟏𝟏𝟏 minutes, 𝟖𝟖𝟏𝟏 𝐋𝐋 will have drained into the bottom

tank after 𝟏𝟏𝟏𝟏 minutes.

That is 𝟖𝟖𝟏𝟏,𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒. To find the height, divide the volume by the area of the base.

𝟖𝟖𝟏𝟏,𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒

𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐦𝐦 ∙ 𝟒𝟒𝟓𝟓 𝐜𝐜𝐦𝐦≈ 𝟐𝟐𝟐𝟐.𝟒𝟒 𝐜𝐜𝐦𝐦

After 𝟏𝟏𝟏𝟏 minutes, the height of the water in the bottom tank will be about 𝟐𝟐𝟒𝟒 𝐜𝐜𝐦𝐦.

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7•3 Lesson 25


𝟔𝟔 𝐜𝐜𝐦𝐦

𝟖𝟖 𝐜𝐜𝐦𝐦 𝟏𝟏𝟒𝟒 𝐜𝐜𝐦𝐦

𝟏𝟏𝟓𝟓 𝐜𝐜𝐦𝐦

𝟓𝟓 𝐜𝐜𝐦𝐦

𝟐𝟐𝟏𝟏 𝐜𝐜𝐦𝐦


𝟖𝟖 𝐜𝐜𝐦𝐦

𝟓𝟓 𝐜𝐜𝐦𝐦 𝟐𝟐𝟐𝟐 𝐜𝐜𝐦𝐦 𝟕𝟕 𝐜𝐜𝐦𝐦

𝟖𝟖 𝐜𝐜𝐦𝐦


𝟏𝟏𝟕𝟕 𝐜𝐜𝐦𝐦

(a) (b)

(c) (d)

6. Two tanks with equal volumes are shown below. The tops are open. The owner wants to cover one tank with a glass top. The cost of glass is $𝟏𝟏.𝟏𝟏𝟓𝟓 per square inch. Which tank would be less expensive to cover? How much less?

Dimensions: 𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. long by 𝟖𝟖 𝐢𝐢𝐢𝐢. wide by 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. high Dimensions: 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢. long by 𝟖𝟖 𝐢𝐢𝐢𝐢. wide by 𝟖𝟖 𝐢𝐢𝐢𝐢. high

Surface area: 𝟒𝟒𝟔𝟔 𝐢𝐢𝐢𝐢𝟐𝟐 Surface area: 𝟏𝟏𝟐𝟐𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐

Cost: $𝟏𝟏.𝟏𝟏𝟓𝟓𝐢𝐢𝐢𝐢𝟐𝟐

∙ 𝟒𝟒𝟔𝟔 𝐢𝐢𝐢𝐢𝟐𝟐 = $𝟒𝟒. 𝟖𝟖𝟏𝟏 Cost: $𝟏𝟏.𝟏𝟏𝟓𝟓𝐢𝐢𝐢𝐢𝟐𝟐

∙ 𝟏𝟏𝟐𝟐𝟏𝟏 𝐢𝐢𝐢𝐢𝟐𝟐 = $𝟔𝟔. 𝟏𝟏𝟏𝟏

The first tank is less expensive. It is $𝟏𝟏.𝟐𝟐𝟏𝟏 cheaper.

7. Each prism below is a gift box sold at the craft store.

a. What is the volume of each prism?

(a) Volume = 𝟒𝟒𝟒𝟒𝟔𝟔 𝐜𝐜𝐦𝐦𝟒𝟒, (b) Volume = 𝟕𝟕𝟓𝟓𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒, (c) Volume = 𝟒𝟒𝟒𝟒𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒, (d) Volume = 𝟏𝟏𝟏𝟏𝟒𝟒𝟏𝟏.𝟓𝟓 𝐜𝐜𝐦𝐦𝟒𝟒

b. Jenny wants to fill each box with jelly beans. If one ounce of jelly beans is approximately 𝟒𝟒𝟏𝟏 𝐜𝐜𝐦𝐦𝟒𝟒, estimate how many ounces of jelly beans Jenny will need to fill all four boxes? Explain your estimates.

Divide each volume in cubic centimeters by 𝟒𝟒𝟏𝟏.

(a) 𝟏𝟏𝟏𝟏.𝟐𝟐 ounces (b) 𝟐𝟐𝟓𝟓 ounces (c) 𝟒𝟒𝟒𝟒 ounces (d) 𝟒𝟒𝟕𝟕.𝟕𝟕 ounces

Jenny would need a total of 𝟏𝟏𝟏𝟏𝟔𝟔.𝟒𝟒 ounces.

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7•3 Lesson 25


8. Two rectangular tanks are filled at a rate of 𝟏𝟏.𝟓𝟓 cubic inches per minute. How long will it take each tank to be half-full?

a. Tank 1 Dimensions: 𝟏𝟏𝟓𝟓 𝐢𝐢𝐢𝐢. by 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. by 𝟏𝟏𝟐𝟐.𝟓𝟓 𝐢𝐢𝐢𝐢.

Volume: 𝟏𝟏,𝟖𝟖𝟕𝟕𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒

Half of the volume is 𝟒𝟒𝟒𝟒𝟕𝟕.𝟓𝟓 𝐢𝐢𝐢𝐢𝟒𝟒.

To find the number of minutes, divide the volume by the rate in cubic inches per minute.

Time: 𝟏𝟏,𝟖𝟖𝟕𝟕𝟓𝟓 minutes.

b. Tank 2 Dimensions: 𝟐𝟐𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. by 𝟒𝟒𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢. by 𝟒𝟒𝟒𝟒𝟖𝟖 𝐢𝐢𝐢𝐢.

Volume: 𝟐𝟐𝟔𝟔𝟐𝟐𝟓𝟓𝟔𝟔𝟒𝟒 𝐢𝐢𝐢𝐢𝟒𝟒

Half of the volume is 𝟐𝟐𝟔𝟔𝟐𝟐𝟓𝟓𝟏𝟏𝟐𝟐𝟖𝟖 𝐢𝐢𝐢𝐢𝟒𝟒.

To find the number of minutes, divide the volume by the rate in cubic inches per minute.

Time: 𝟒𝟒𝟏𝟏 minutes

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7•3 Lesson 26


5 cm

3 cm

2 cm

6 cm


Student Outcomes

Students solve real-world and mathematical problems involving volume and surface areas of three-dimensional objects composed of cubes and right prisms.

Lesson Notes In this lesson, students apply what they learned in Lessons 22–25 to solve real-world problems. As students work the problems, encourage them to present their approaches for determining volume and surface area. Students use volume formulas to find the volume of a right prism that has had part of its volume displaced by another prism. Students work with cubic units and units of liquid measure on the volume problems. Students also continue to calculate surface area.

Classwork

Opening (2 minutes)

In the Opening Exercise, students are asked to find the area of a region obtained by cutting a smaller rectangle out of the middle of a larger rectangle. This exercise provides information about students who may need some additional support during the lesson if they have difficulty solving this problem. Tell the class that today they are applying what they learned about finding the surface area and volume of prisms to real-world problems.


Explain to your partner how you would calculate the area of the shaded region. Then, calculate the area.

Find the area of the outer rectangle, and subtract the area of the inner rectangle.

𝟔𝟔 𝐜𝐜𝐜𝐜 × 𝟑𝟑 𝐜𝐜𝐜𝐜− 𝟓𝟓 𝐜𝐜𝐜𝐜 × 𝟐𝟐 𝐜𝐜𝐜𝐜 = 𝟖𝟖 𝐜𝐜𝐜𝐜𝟐𝟐

Scaffolding: If students are struggling, have them actually cut out the figures and take measurements.

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7•3 Lesson 26


Example 1 (6 minutes): Volume of a Shell

This example builds on the area problem in the Opening Exercise, but extends to volume by cutting a smaller rectangular prism out of a larger cube to form an insulated box. Depending on the level of students, guide them through this example, allow them to work with a partner, or allow them to work in small groups. If students work with a partner or a group, be sure to present different solutions and monitor the groups’ progress.

Example 1

The insulated box shown is made from a large cube with a hollow inside that is a right rectangular prism with a square base. The figure on the right is what the box looks like from above.

a. Calculate the volume of the outer box.

𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜 × 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜 × 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜 = 𝟏𝟏𝟑𝟑,𝟖𝟖𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜𝟑𝟑

b. Calculate the volume of the inner prism.

𝟏𝟏𝟖𝟖 𝐜𝐜𝐜𝐜 × 𝟏𝟏𝟖𝟖 𝐜𝐜𝐜𝐜 × 𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜 = 𝟔𝟔,𝟖𝟖𝟖𝟖𝟓𝟓 𝐜𝐜𝐜𝐜𝟑𝟑

c. Describe in words how you would find the volume of the insulation.

Find the volume of the outer cube and the inner right rectangular prism, and then subtract the two volumes.

d. Calculate the volume of the insulation in cubic centimeters.

𝟏𝟏𝟑𝟑,𝟖𝟖𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜𝟑𝟑 − 𝟔𝟔,𝟖𝟖𝟖𝟖𝟓𝟓 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟔𝟔,𝟗𝟗𝟑𝟑𝟗𝟗𝐜𝐜𝐜𝐜𝟑𝟑

e. Calculate the amount of water the box can hold in liters.

𝟔𝟔𝟗𝟗𝟑𝟑𝟗𝟗 𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟔𝟔𝟗𝟗𝟑𝟑𝟗𝟗 𝐜𝐜𝐦𝐦 =(𝟔𝟔𝟗𝟗𝟑𝟑𝟗𝟗 𝐜𝐜𝐦𝐦)

𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝐜𝐜𝐦𝐦𝐦𝐦 = 𝟔𝟔.𝟗𝟗𝟑𝟑𝟗𝟗 𝐦𝐦

Use these questions with the whole class or small groups as discussion points.

How did you calculate the volume of the insulation?

First, calculate the volume of the outer cube, and then subtract the volume of the inner prism.

How do you convert cubic centimeters to liters?

1 cm3 = 1 mL and 1,000 mL = 1 L, so divide by 1,000.

Top View

Insulation

18 cm

24 cm 24 cm 18 cm

21 14 cm

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7•3 Lesson 26


Exercise 1 (15 minutes): Brick Planter Design

In this exercise, students construct a brick planter and determine the amount of soil the planter will hold. First, students calculate the number of bricks needed to build the planter; then, they determine the cost of building the planter and filling it with soil. Have the class consider these questions as they discuss this exercise.

How do you determine the internal dimensions?

Find the external dimensions, and subtract the thickness of the shell.

Explain how to determine the number of bricks needed. If you were going to construct this planter, can you think of other factors that would have to be considered?

Find the volume of the bricks and divide by the volume of one brick. Other factors to consider include whether the bricks are perfectly rectangular and whether or not grout is used.

What do you think about your calculated cost? Is it what you expected? If not, is it higher or lower than you expected? Why?

This is a very open-ended question, and answers will depend on what the students originally thought. Some will say the cost is higher, while others will say it is lower.

Exercise 1: Brick Planter Design

You have been asked by your school to design a brick planter that will be used by classes to plant flowers. The planter will be built in the shape of a right rectangular prism with no bottom so water and roots can access the ground beneath.

The exterior dimensions are to be 𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. × 𝟗𝟗 𝐟𝐟𝐟𝐟. × 𝟐𝟐𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. The bricks used to construct the planter are 𝟔𝟔 𝐢𝐢𝐢𝐢. long, 𝟑𝟑𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. wide, and 𝟐𝟐 𝐢𝐢𝐢𝐢. high.

a. What are the interior dimensions of the planter if the thickness of the planter’s walls is equal to the length of the bricks?

𝟔𝟔 𝐢𝐢𝐢𝐢 = 𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.

Interior length:

𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.−𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.−𝟏𝟏

𝟐𝟐 𝐟𝐟𝐟𝐟. = 𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟.

Interior width:

𝟗𝟗 𝐟𝐟𝐟𝐟.−𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.−𝟏𝟏

𝟐𝟐 𝐟𝐟𝐟𝐟. = 𝟖𝟖 𝐟𝐟𝐟𝐟.

Interior dimensions:

𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟.× 𝟖𝟖 𝐟𝐟𝐟𝐟.× 𝟐𝟐𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.

b. What is the volume of the bricks that form the planter?

Solution 1

Subtract the volume of the smaller interior prism 𝑽𝑽𝑺𝑺 from the volume of the large exterior prism 𝑽𝑽𝑳𝑳.

𝑽𝑽𝑩𝑩𝑩𝑩𝑩𝑩𝑩𝑩𝑩𝑩 = 𝑽𝑽𝑳𝑳 − 𝑽𝑽𝑺𝑺

𝑽𝑽𝐁𝐁𝐁𝐁𝐢𝐢𝐜𝐜𝐁𝐁 = �𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.× 𝟗𝟗 𝐟𝐟𝐟𝐟. × 𝟐𝟐𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. � − �𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟.× 𝟖𝟖 𝐟𝐟𝐟𝐟.× 𝟐𝟐𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. �

𝑽𝑽𝐁𝐁𝐁𝐁𝐢𝐢𝐜𝐜𝐁𝐁 = 𝟐𝟐𝟐𝟐𝟏𝟏 𝐟𝐟𝐟𝐟𝟑𝟑 − 𝟐𝟐𝟐𝟐𝟏𝟏 𝐟𝐟𝐟𝐟𝟑𝟑

𝑽𝑽𝐁𝐁𝐁𝐁𝐢𝐢𝐜𝐜𝐁𝐁 = 𝟓𝟓𝟏𝟏 𝐟𝐟𝐟𝐟𝟑𝟑

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7•3 Lesson 26


Solution 2

The volume of the brick is equal to the area of the base times the height.

𝑩𝑩 = 𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.× �𝟖𝟖𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. +𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. +𝟖𝟖𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. +𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. �

𝑩𝑩 = 𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.× (𝟐𝟐𝟏𝟏 𝐟𝐟𝐟𝐟. ) = 𝟐𝟐𝟏𝟏 𝐟𝐟𝐟𝐟𝟐𝟐


𝑽𝑽 = (𝟐𝟐𝟏𝟏 𝐟𝐟𝐟𝐟𝟐𝟐) �𝟐𝟐𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. � = 𝟓𝟓𝟏𝟏 𝐟𝐟𝐟𝐟𝟑𝟑

c. If you are going to fill the planter 𝟑𝟑𝟐𝟐 full of soil, how much soil will you need to purchase, and what will be the height of the soil?

The height of the soil will be 𝟑𝟑𝟐𝟐 of 𝟐𝟐𝟏𝟏𝟐𝟐 feet.

𝟑𝟑𝟐𝟐

�𝟓𝟓𝟐𝟐𝐟𝐟𝐟𝐟. � =

𝟏𝟏𝟓𝟓𝟖𝟖

𝐟𝐟𝐟𝐟.; The height of the soil will be 𝟏𝟏𝟓𝟓𝟖𝟖 𝐟𝐟𝐟𝐟. (or 𝟏𝟏𝟐𝟐𝟖𝟖 𝐟𝐟𝐟𝐟.).

The volume of the soil in the planter:

𝑽𝑽 = �𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟× 𝟖𝟖 𝐟𝐟𝐟𝐟.× 𝟏𝟏𝟓𝟓𝟖𝟖 𝐟𝐟𝐟𝐟. �

𝑽𝑽 = (𝟏𝟏𝟏𝟏 𝐟𝐟𝐟𝐟.× 𝟏𝟏𝟓𝟓 𝐟𝐟𝐟𝐟𝟐𝟐) = 𝟏𝟏𝟔𝟔𝟓𝟓 𝐟𝐟𝐟𝐟𝟑𝟑

d. How many bricks are needed to construct the planter?

𝑷𝑷 = 𝟐𝟐�𝟖𝟖𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. � + 𝟐𝟐�𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. �

𝑷𝑷 = 𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. +𝟐𝟐𝟑𝟑 𝐟𝐟𝐟𝐟. = 𝟐𝟐𝟏𝟏 𝐟𝐟𝐟𝐟.

𝟑𝟑𝟏𝟏𝟐𝟐 𝐢𝐢𝐢𝐢. = 𝟐𝟐𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟.

We can then divide the perimeter by the width of each brick in order to determine the number of bricks needed for each layer of the planter.

𝟐𝟐𝟏𝟏÷𝟐𝟐𝟐𝟐𝟐𝟐

=𝟗𝟗𝟔𝟔𝟏𝟏𝟐𝟐

𝟐𝟐𝟏𝟏×𝟐𝟐𝟐𝟐𝟐𝟐

=𝟗𝟗𝟔𝟔𝟏𝟏𝟐𝟐

≈ 𝟏𝟏𝟑𝟑𝟐𝟐.𝟏𝟏

Each layer of the planter requires approximately 𝟏𝟏𝟑𝟑𝟐𝟐.𝟏𝟏 bricks.

Scaffolding: Solution 2 is an extension of thinking from Lesson 3 Example 6, in which students found various ways to write expressions representing a tiled perimeter around a rectangle.

Scaffolding: If you have students who need a challenge, have them extend this concept to the volume of a cylinder and the volume of the metal part of a can.

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7•3 Lesson 26


𝟐𝟐 𝐢𝐢𝐢𝐢. = 𝟏𝟏𝟔𝟔 𝐟𝐟𝐟𝐟.

The height of the planter, 𝟐𝟐𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟., is equal to the product of the number of layers of brick, 𝒏𝒏, and the height of

each brick, 𝟏𝟏𝟔𝟔 𝐟𝐟𝐟𝐟.

𝟐𝟐𝟏𝟏𝟐𝟐

= 𝒍𝒍 �𝟏𝟏𝟔𝟔

�

𝟔𝟔�𝟐𝟐𝟏𝟏𝟐𝟐� = 𝒍𝒍

𝟏𝟏𝟓𝟓 = 𝒍𝒍

There are 𝟏𝟏𝟓𝟓 layers of bricks in the planter.

The total number of bricks, 𝒃𝒃, is equal to the product of the number of bricks in each layer �𝟗𝟗𝟔𝟔𝟏𝟏𝟐𝟐 � and the number of layers (𝟏𝟏𝟓𝟓).

𝒃𝒃 = �𝟗𝟗𝟔𝟔𝟏𝟏𝟐𝟐� (𝟏𝟏𝟓𝟓)

𝒃𝒃 =𝟏𝟏𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐

≈ 𝟐𝟐𝟏𝟏𝟓𝟓𝟐𝟐.𝟏𝟏

It is not reasonable to purchase 𝟏𝟏.𝟏𝟏 brick; we must round up to the next whole brick, which is 𝟐𝟐,𝟏𝟏𝟓𝟓𝟖𝟖 bricks. Therefore, 𝟐𝟐,𝟏𝟏𝟓𝟓𝟖𝟖 bricks are needed to construct the planter.

e. Each brick used in this project costs $𝟏𝟏.𝟖𝟖𝟐𝟐 and weighs 𝟐𝟐.𝟓𝟓 𝐥𝐥𝐥𝐥. The supply company charges a delivery fee of $𝟏𝟏𝟓𝟓 per whole ton (𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥) over 𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥 How much will your school pay for the bricks (including delivery) to construct the planter?

If the school purchases 𝟐𝟐𝟏𝟏𝟓𝟓𝟖𝟖 bricks, the total weight of the bricks for the planter,

𝟐𝟐𝟏𝟏𝟓𝟓𝟖𝟖(𝟐𝟐.𝟓𝟓 𝐥𝐥𝐥𝐥) = 𝟗𝟗𝟐𝟐𝟔𝟔𝟏𝟏 𝐥𝐥𝐥𝐥

The number of whole tons over 𝟐𝟐,𝟏𝟏𝟏𝟏𝟏𝟏 pounds,

𝟗𝟗𝟐𝟐𝟔𝟔𝟏𝟏 − 𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟓𝟓𝟐𝟐𝟔𝟔𝟏𝟏

Since 𝟏𝟏 𝐟𝐟𝐭𝐭𝐢𝐢 = 𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥., there are 𝟐𝟐 whole tons (𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥.) in 𝟓𝟓,𝟐𝟐𝟔𝟔𝟏𝟏 𝐥𝐥𝐥𝐥.

Total cost = cost of bricks + cost of delivery

Total cost = 𝟏𝟏.𝟖𝟖𝟐𝟐(𝟐𝟐𝟏𝟏𝟓𝟓𝟖𝟖) + 𝟐𝟐(𝟏𝟏𝟓𝟓)

Total cost = 𝟏𝟏𝟔𝟔𝟖𝟖𝟐𝟐.𝟓𝟓𝟔𝟔 + 𝟑𝟑𝟏𝟏 = 𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐.𝟓𝟓𝟔𝟔

The cost for bricks and delivery will be $𝟏𝟏,𝟐𝟐𝟏𝟏𝟐𝟐.𝟓𝟓𝟔𝟔.

f. A cubic foot of topsoil weighs between 𝟐𝟐𝟓𝟓 and 𝟏𝟏𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥. How much will the soil in the planter weigh?

The volume of the soil in the planter is 𝟏𝟏𝟔𝟔𝟓𝟓 𝐟𝐟𝐟𝐟𝟑𝟑.

Minimum weight: Maximum weight:

Minimum weight = 𝟐𝟐𝟓𝟓 𝐥𝐥𝐥𝐥(𝟏𝟏𝟔𝟔𝟓𝟓) Maximum weight = 𝟏𝟏𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥(𝟏𝟏𝟔𝟔𝟓𝟓)

Minimum weight = 𝟏𝟏𝟐𝟐𝟑𝟑𝟐𝟐𝟓𝟓 𝐥𝐥𝐥𝐥. Maximum weight = 𝟏𝟏𝟔𝟔𝟓𝟓𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥. The soil in the planter will weigh between 𝟏𝟏𝟐𝟐,𝟑𝟑𝟐𝟐𝟓𝟓 𝐥𝐥𝐥𝐥. and 𝟏𝟏𝟔𝟔,𝟓𝟓𝟏𝟏𝟏𝟏 𝐥𝐥𝐥𝐥.

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g. If the topsoil costs $𝟏𝟏.𝟖𝟖𝟖𝟖 per cubic foot, calculate the total cost of materials that will be used to construct the planter.

The total cost of the top soil:

Cost = 𝟏𝟏.𝟖𝟖𝟖𝟖(𝟏𝟏𝟔𝟔𝟓𝟓) = 𝟏𝟏𝟐𝟐𝟓𝟓.𝟐𝟐; The cost of the top soil will be $𝟏𝟏𝟐𝟐𝟓𝟓.𝟐𝟐𝟏𝟏.

The total cost of materials for the brick planter project:

Cost = (cost of bricks) + (cost of soil)

Cost = $𝟏𝟏,𝟐𝟐𝟏𝟏𝟐𝟐.𝟓𝟓𝟔𝟔+ $𝟏𝟏𝟐𝟐𝟓𝟓.𝟐𝟐𝟏𝟏

Cost = $𝟏𝟏,𝟖𝟖𝟔𝟔𝟐𝟐.𝟐𝟐𝟔𝟔

The total cost of materials for the brick planter project will be $𝟏𝟏,𝟖𝟖𝟔𝟔𝟐𝟐.𝟐𝟐𝟔𝟔.

Exercise 2 (12 minutes): Design a Feeder

This is a very open-ended task. If students struggled with the first example and exercise, consider moving them directly to some of the Problem Set exercises. Students may at first struggle to determine a figure that will work, but refer back to some of the designs from earlier lessons. Right prisms with triangular bases or trapezoidal bases would work well. Encourage students to find reasonable dimensions and to be sure the volume is in the specified range. Students may try various approaches to this problem. Students may work with partners or in groups, but be sure to bring the class back together, and have students present their designs and costs. Discuss the pros and cons of each design. Consider having a contest for the best design.

Exercise 2: Design a Feeder

You did such a good job designing the planter that a local farmer has asked you to design a feeder for the animals on his farm. Your feeder must be able to contain at least 𝟏𝟏𝟏𝟏𝟏𝟏,𝟏𝟏𝟏𝟏𝟏𝟏 cubic centimeters, but not more than 𝟐𝟐𝟏𝟏𝟏𝟏,𝟏𝟏𝟏𝟏𝟏𝟏 cubic centimeters of grain when it is full. The feeder is to be built of stainless steel and must be in the shape of a right prism but not a right rectangular prism. Sketch your design below including dimensions. Calculate the volume of grain that it can hold and the amount of metal needed to construct the feeder.

The farmer needs a cost estimate. Calculate the cost of constructing the feeder if 𝟏𝟏𝟐𝟐𝐜𝐜𝐜𝐜 thick stainless steel sells for

$𝟗𝟗𝟑𝟑.𝟐𝟐𝟓𝟓 per square meter.

Answers will vary. Below is an example using a right trapezoidal prism.

This feeder design consists of an open-top container in the shape of a right trapezoidal prism. The trapezoidal sides of the feeder will allow animals easier access to feed at its bottom. The dimensions of the feeder are shown in the diagram.

𝑩𝑩 =𝟏𝟏𝟐𝟐

(𝒃𝒃𝟏𝟏 + 𝒃𝒃𝟐𝟐)𝑩𝑩

𝑩𝑩 =𝟏𝟏𝟐𝟐

(𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 + 𝟖𝟖𝟏𝟏 𝐜𝐜𝐜𝐜) ∙ 𝟑𝟑𝟏𝟏 𝐜𝐜𝐜𝐜

𝑩𝑩 =𝟏𝟏𝟐𝟐

(𝟏𝟏𝟖𝟖𝟏𝟏 𝐜𝐜𝐜𝐜) ∙ 𝟑𝟑𝟏𝟏 𝐜𝐜𝐜𝐜

𝑩𝑩 = 𝟗𝟗𝟏𝟏 𝐜𝐜𝐜𝐜 ∙ 𝟑𝟑𝟏𝟏 𝐜𝐜𝐜𝐜

𝑩𝑩 = 𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

𝑽𝑽 = 𝑩𝑩𝑩𝑩 𝑽𝑽 = (𝟐𝟐,𝟐𝟐𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐)(𝟔𝟔𝟏𝟏 𝐜𝐜𝐜𝐜) 𝑽𝑽 = 𝟏𝟏𝟔𝟔𝟐𝟐,𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑

The volume of the solid prism is 𝟏𝟏𝟔𝟔𝟐𝟐,𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑, so the volume that the feeder can contain is slightly less, depending on the thickness of the metal used.

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The exterior surface area of the feeder tells us the area of metal required to build the feeder.

𝑺𝑺𝑺𝑺 = �𝑳𝑳𝑺𝑺 − 𝑺𝑺𝒕𝒕𝒕𝒕𝒕𝒕�+ 𝟐𝟐𝑩𝑩

𝑺𝑺𝑺𝑺 = 𝟔𝟔𝟏𝟏 𝐜𝐜𝐜𝐜 ∙ (𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜+ 𝟖𝟖𝟏𝟏 𝐜𝐜𝐜𝐜 + 𝟐𝟐𝟏𝟏 𝐜𝐜𝐜𝐜) + 𝟐𝟐(𝟐𝟐,𝟐𝟐𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐)

𝑺𝑺𝑺𝑺 = 𝟔𝟔𝟏𝟏 𝐜𝐜𝐜𝐜(𝟏𝟏𝟔𝟔𝟏𝟏 𝐜𝐜𝐜𝐜) + 𝟓𝟓,𝟐𝟐𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

𝑺𝑺𝑺𝑺 = 𝟗𝟗,𝟔𝟔𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 + 𝟓𝟓,𝟐𝟐𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

𝑺𝑺𝑺𝑺 = 𝟏𝟏𝟓𝟓,𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐

The feeder will require 𝟏𝟏𝟓𝟓,𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 of metal.

𝟏𝟏 𝐜𝐜𝟐𝟐 = 𝟏𝟏𝟏𝟏,𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐, so 𝟏𝟏𝟓𝟓,𝟏𝟏𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜𝟐𝟐 = 𝟏𝟏.𝟓𝟓 𝐜𝐜𝟐𝟐

Cost = 𝟗𝟗𝟑𝟑.𝟐𝟐𝟓𝟓(𝟏𝟏.𝟓𝟓) = 𝟏𝟏𝟑𝟑𝟗𝟗.𝟖𝟖𝟐𝟐𝟓𝟓

Since this is a measure of money, the cost must be rounded to the nearest cent, which is $𝟏𝟏𝟑𝟑𝟗𝟗.𝟖𝟖𝟖𝟖.

Closing (2 minutes)

Describe the process of finding the volume of a prism shell.

Find the volume of the outer figure; then, subtract the volume of the inner figure. How does the thickness of the shell affect the internal dimensions of the prism? The internal volume? The

external volume?

The thicker the shell, the smaller the internal dimensions, and the smaller the internal volume. The external volume is not affected by the thickness of the shell.


The Exit Ticket problem includes scaffolding. If students grasp this concept well, assign only parts (c) and (d) of the Exit Ticket.

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Name Date


Exit Ticket Lawrence is designing a cooling tank that is a square prism. A pipe in the shape of a smaller 2 ft × 2 ft square prism passes through the center of the tank as shown in the diagram, through which a coolant will flow.

a. What is the volume of the tank including the cooling pipe?

b. What is the volume of coolant that fits inside the cooling pipe?

c. What is the volume of the shell (the tank not including the cooling pipe)?

d. Find the surface area of the cooling pipe.

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Lawrence is designing a cooling tank that is a square prism. A pipe in the shape of a smaller 𝟐𝟐 𝐟𝐟𝐟𝐟× 𝟐𝟐 𝐟𝐟𝐟𝐟 square prism passes through the center of the tank as shown in the diagram, through which a coolant will flow.

a. What is the volume of the tank including the cooling pipe?

𝟐𝟐 𝐟𝐟𝐟𝐟.× 𝟑𝟑 𝐟𝐟𝐟𝐟.× 𝟑𝟑 𝐟𝐟𝐟𝐟. = 𝟔𝟔𝟑𝟑 𝐟𝐟𝐟𝐟𝟑𝟑

b. What is the volume of coolant that fits inside the cooling pipe?

𝟐𝟐 𝐟𝐟𝐟𝐟.× 𝟐𝟐 𝐟𝐟𝐟𝐟.× 𝟐𝟐 𝐟𝐟𝐟𝐟. = 𝟐𝟐𝟖𝟖 𝐟𝐟𝐟𝐟𝟑𝟑

c. What is the volume of the shell (the tank not including the cooling pipe)?

𝟔𝟔𝟑𝟑 𝐟𝐟𝐟𝐟𝟑𝟑 − 𝟐𝟐𝟖𝟖 𝐟𝐟𝐟𝐟𝟑𝟑 = 𝟑𝟑𝟓𝟓 𝐟𝐟𝐟𝐟𝟑𝟑

d. Find the surface area of the cooling pipe.

𝟐𝟐 𝐟𝐟𝐟𝐟.× 𝟐𝟐 𝐟𝐟𝐟𝐟.× 𝟐𝟐 = 𝟓𝟓𝟔𝟔 𝐟𝐟𝐟𝐟𝟐𝟐


1. A child’s toy is constructed by cutting a right triangular prism out of a right rectangular prism.

a. Calculate the volume of the rectangular prism.

𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 × 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 × 𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜 = 𝟏𝟏𝟐𝟐𝟓𝟓𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑

b. Calculate the volume of the triangular prism.

𝟏𝟏𝟐𝟐�𝟓𝟓 𝐜𝐜𝐜𝐜 × 𝟐𝟐

𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜�× 𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜 = 𝟐𝟐𝟖𝟖𝟏𝟏𝟖𝟖


c. Calculate the volume of the material remaining in the rectangular prism.

𝟏𝟏𝟐𝟐𝟓𝟓𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑 − 𝟐𝟐𝟖𝟖 𝟏𝟏𝟖𝟖

𝐜𝐜𝐜𝐜𝟑𝟑 = 𝟏𝟏𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐𝟖𝟖


12 ½ cm

10 cm

10 cm

Top View

5 cm 2 ½ cm

5.6 cm

Scaffolding: If students have mastered this concept easily, assign only parts (c) and (d).

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d. What is the largest number of triangular prisms that can be cut from the rectangular prism?

𝟏𝟏𝟐𝟐𝟓𝟓𝟏𝟏 𝐜𝐜𝐜𝐜𝟑𝟑

𝟐𝟐𝟖𝟖 𝟏𝟏𝟖𝟖 𝐜𝐜𝐜𝐜𝟑𝟑= 𝟏𝟏𝟔𝟔

e. What is the surface area of the triangular prism (assume there is no top or bottom)?

𝟓𝟓.𝟔𝟔 𝐜𝐜𝐜𝐜 × 𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜 + 𝟐𝟐𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜 × 𝟏𝟏𝟐𝟐𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜 + 𝟓𝟓 𝐜𝐜𝐜𝐜 × 𝟏𝟏𝟐𝟐 𝟏𝟏𝟐𝟐

𝐜𝐜𝐜𝐜 = 𝟏𝟏𝟔𝟔𝟑𝟑𝟑𝟑𝟐𝟐


2. A landscape designer is constructing a flower bed in the shape of a right trapezoidal prism. He needs to run three identical square prisms through the bed for drainage.

a. What is the volume of the bed without the drainage pipes?

𝟏𝟏𝟐𝟐

(𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. +𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. ) × 𝟑𝟑 𝐟𝐟𝐟𝐟.× 𝟏𝟏𝟔𝟔 𝐟𝐟𝐟𝐟. = 𝟔𝟔𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟𝟑𝟑

b. What is the total volume of the three drainage pipes?

𝟑𝟑�𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟𝟐𝟐 × 𝟏𝟏𝟔𝟔 𝐟𝐟𝐟𝐟. � = 𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟𝟑𝟑

c. What is the volume of soil if the planter is filled to 𝟑𝟑𝟐𝟐

of its total capacity with the pipes in place?

𝟑𝟑𝟐𝟐

(𝟔𝟔𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟𝟑𝟑) − 𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟𝟑𝟑 = 𝟐𝟐𝟓𝟓𝟔𝟔 𝐟𝐟𝐟𝐟𝟑𝟑

d. What is the height of the soil? If necessary, round to the nearest tenth.

𝟐𝟐𝟓𝟓𝟔𝟔 𝐟𝐟𝐟𝐟𝟑𝟑

𝟏𝟏𝟐𝟐 (𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. +𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. ) × 𝟏𝟏𝟔𝟔 𝐟𝐟𝐟𝐟.

≈ 𝟐𝟐.𝟐𝟐 𝐟𝐟𝐟𝐟.

e. If the bed is made of 𝟖𝟖 𝐟𝐟𝐟𝐟. × 𝟐𝟐 𝐟𝐟𝐟𝐟. pieces of plywood, how many pieces of plywood will the landscape designer need to construct the bed without the drainage pipes?

𝟐𝟐�𝟑𝟑𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟.× 𝟏𝟏𝟔𝟔 𝐟𝐟𝐟𝐟. �+ 𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.× 𝟏𝟏𝟔𝟔 𝐟𝐟𝐟𝐟. +𝟐𝟐�𝟏𝟏𝟐𝟐

(𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. +𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟. ) × 𝟑𝟑 𝐟𝐟𝐟𝐟. � = 𝟑𝟑𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟𝟐𝟐

𝟑𝟑𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟𝟐𝟐 ÷(𝟖𝟖 𝐟𝐟𝐟𝐟.× 𝟐𝟐 𝐟𝐟𝐟𝐟. )

𝐩𝐩𝐢𝐢𝐩𝐩𝐜𝐜𝐩𝐩 𝐭𝐭𝐟𝐟 𝐩𝐩𝐥𝐥𝐩𝐩𝐩𝐩𝐭𝐭𝐭𝐭𝐩𝐩= 𝟏𝟏𝟏𝟏.𝟐𝟐, or 𝟏𝟏𝟐𝟐 pieces of plywood

f. If the plywood needed to construct the bed costs $𝟑𝟑𝟓𝟓 per 𝟖𝟖 𝐟𝐟𝐟𝐟.× 𝟐𝟐 𝐟𝐟𝐟𝐟. piece, the drainage pipes cost $𝟏𝟏𝟐𝟐𝟓𝟓 each, and the soil costs $𝟏𝟏.𝟐𝟐𝟓𝟓/cubic foot, how much does it cost to construct and fill the bed?

$𝟑𝟑𝟓𝟓𝐩𝐩𝐢𝐢𝐩𝐩𝐜𝐜𝐩𝐩 𝐭𝐭𝐟𝐟 𝐩𝐩𝐥𝐥𝐩𝐩𝐩𝐩𝐭𝐭𝐭𝐭𝐩𝐩

(𝟏𝟏𝟐𝟐 𝐩𝐩𝐢𝐢𝐩𝐩𝐜𝐜𝐩𝐩𝐩𝐩 𝐭𝐭𝐟𝐟 𝐩𝐩𝐥𝐥𝐩𝐩𝐩𝐩𝐭𝐭𝐭𝐭𝐩𝐩) +$𝟏𝟏𝟐𝟐𝟓𝟓𝐩𝐩𝐢𝐢𝐩𝐩𝐩𝐩

(𝟑𝟑 𝐩𝐩𝐢𝐢𝐩𝐩𝐩𝐩𝐩𝐩) +$𝟏𝟏.𝟐𝟐𝟓𝟓𝐟𝐟𝐟𝐟𝟑𝟑 𝐩𝐩𝐭𝐭𝐢𝐢𝐥𝐥

(𝟐𝟐𝟓𝟓𝟔𝟔 𝐟𝐟𝐟𝐟𝟑𝟑 𝐩𝐩𝐭𝐭𝐢𝐢𝐥𝐥) = $𝟏𝟏,𝟑𝟑𝟔𝟔𝟓𝟓.𝟏𝟏𝟏𝟏

Side View

𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.

𝟏𝟏𝟐𝟐 𝐟𝐟𝐟𝐟.

𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟𝟐𝟐

𝟑𝟑 𝐟𝐟𝐟𝐟.

𝟏𝟏𝟔𝟔 𝐟𝐟𝐟𝐟. 𝟑𝟑𝟏𝟏𝟐𝟐

𝐟𝐟𝐟𝐟

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7•3 End-of-Module Assessment Task

Name Date

1. Gloria says the two expressions 14

(12𝑥𝑥 + 24) − 9𝑥𝑥 and −6(𝑥𝑥 + 1) are equivalent. Is she correct? Explain how you know.

2. A grocery store has advertised a sale on ice cream. Each carton of any flavor of ice cream costs $3.79.

a. If Millie buys one carton of strawberry ice cream and one carton of chocolate ice cream, write an

algebraic expression that represents the total cost of buying the ice cream.

b. Write an equivalent expression for your answer in part (a).

c. Explain how the expressions are equivalent.

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3. A new park was designed to contain two circular gardens. Garden A has a diameter of 50 m, and garden B has a diameter of 70 m. a. If the gardener wants to outline the gardens in edging, how many meters will be needed to outline

the smaller garden? (Write in terms of 𝜋𝜋.)

b. How much more edging will be needed for the larger garden than the smaller one? (Write in terms of 𝜋𝜋.)

c. The gardener wishes to put down weed block fabric on the two gardens before the plants are planted in the ground. How much fabric will be needed to cover the area of both gardens? (Write in terms of 𝜋𝜋.)

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4. A play court on the school playground is shaped like a square joined by a semicircle. The perimeter around the entire play court is 182.8 ft., and 62.8 ft. of the total perimeter comes from the semicircle.

a. What is the radius of the semicircle? Use 3.14 for 𝜋𝜋.

b. The school wants to cover the play court with sports court flooring. Using 3.14 for 𝜋𝜋, how many square feet of flooring does the school need to purchase to cover the play court?

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5. Marcus drew two adjacent angles.

a. If ∠A𝐵𝐵𝐵𝐵 has a measure one-third of ∠𝐵𝐵𝐵𝐵𝐶𝐶, then what is the degree measurement of ∠𝐵𝐵𝐵𝐵𝐶𝐶?

b. If the measure of ∠𝐵𝐵𝐵𝐵𝐶𝐶 is 9(8𝑥𝑥 + 11) degrees, then what is the value of 𝑥𝑥? 6. The dimensions of an above-ground, rectangular pool are 25 feet long, 18 feet wide, and 6 feet deep.

a. How much water is needed to fill the pool?

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b. If there are 7.48 gallons in 1 cubic foot, how many gallons are needed to fill the pool?

c. Assume there was a hole in the pool, and 3,366 gallons of water leaked from the pool. How many feet did the water level drop?

d. After the leak was repaired, it was necessary to lay a thin layer of concrete to protect the sides of the pool. Calculate the area to be covered to complete the job.

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7. Gary is learning about mosaics in art class. His teacher passes out small square tiles and encourages the students to cut up the tiles in various angles. Gary’s first cut tile looks like this:

a. Write an equation relating ∠𝑇𝑇𝑇𝑇𝑇𝑇 with ∠𝑇𝑇𝑇𝑇𝐿𝐿.

b. Solve for 𝑚𝑚.

c. What is the measure of ∠𝑇𝑇𝑇𝑇𝑇𝑇?

d. What is the measure of ∠𝑇𝑇𝑇𝑇𝐿𝐿?

3𝑚𝑚° (𝑚𝑚 − 10)°

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A Progression Toward Mastery

Assessment Task Item

STEP 1 Missing or incorrect answer and little evidence of reasoning or application of mathematics to solve the problem

STEP 2 Missing or incorrect answer but evidence of some reasoning or application of mathematics to solve the problem

STEP 3 A correct answer with some evidence of reasoning or application of mathematics to solve the problem, or an incorrect answer with substantial evidence of solid reasoning or application of mathematics to solve the problem

STEP 4 A correct answer supported by substantial evidence of solid reasoning or application of mathematics to solve the problem

1

7.EE.A.1

Student demonstrates a limited understanding of writing expressions in standard form and determining if they are equivalent expressions. Student shows some knowledge of the distributive property.

Student makes a conceptual error in writing one of the expressions in standard form but writes the other expression correctly and provides an appropriate answer and explanation.

Student writes each expression in correct standard form, −6𝑥𝑥 + 6 and −6𝑥𝑥 − 6. Student indicates the expressions are not equivalent, but no explanation is provided, or the explanation is incorrect. OR Student demonstrates a solid understanding but makes one computational error, such as writing −6(𝑥𝑥 + 1) as −6𝑥𝑥 + 6, and indicates the expressions are not equivalent.

Student writes each expression in correct standard form, −6𝑥𝑥 + 6 and −6𝑥𝑥 − 6. Student indicates the expressions are not equivalent and provides an appropriate explanation.

2 7.EE.A.2 Student work shows little evidence of correct reasoning, such as 𝑠𝑠 + 𝑐𝑐, but no further work is shown or is incorrect. OR Student does not demonstrate an understanding of the meaning of writing equivalent expressions.

Student makes a conceptual error such as distributing or factoring incorrectly.

Student writes a correct algebraic expression for part (a) and an equivalent expression for part (b), but the explanation for part (c) is incorrect or not shown.

Student writes a correct algebraic expression to represent the total cost of two flavors of ice cream, such as $3.79(𝑠𝑠 + 𝑐𝑐). Student writes an equivalent expression for part (a), such as $3.79𝑠𝑠 + $3.79𝑐𝑐, providing an appropriate explanation on how the expressions are equivalent, such as

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applying the distributive property. Sample answers given for parts (a) and (b) could be reversed, and the explanation would include factoring the expression.

3

a

7.G.B.4

Student demonstrates knowing circumference is used, but no further correct work is shown.

Student makes a conceptual error, such as finding the area of the smaller garden correctly, as 625𝜋𝜋. OR Student uses the circumference formula, 𝐵𝐵 = 2𝜋𝜋𝜋𝜋, but uses the diameter instead of the radius to get an answer of 100𝜋𝜋 m. OR Student makes two or more computational and/or labeling errors.

Student recognizes that the concept of circumference must be used, but makes a computational or labeling error. OR Student finds the correct circumference but does not leave the answer in terms of 𝜋𝜋, and instead uses 3.14, getting an answer of 157 m.

Student finds the circumference of the smaller garden correctly as 50𝜋𝜋 m.

b

7.G.B.4

Student demonstrates knowing circumference, but no further correct work is shown.

Student makes a conceptual error, such as incorrectly finding the area of the larger garden as 1225𝜋𝜋. OR Student finds the circumference, using the diameter instead of the radius, and gets 140𝜋𝜋 m. OR Student makes two or more computational and/or labeling errors.

Student recognizes that the concept of circumference must be used but makes a computational or labeling error. OR Student finds the circumference of the larger garden correctly as 70𝜋𝜋 m but does not find how much more fencing is needed for the larger garden compared to the smaller garden, or finds it incorrectly. OR Student finds the correct circumference but does not leave the answer in terms of 𝜋𝜋, and instead uses 3.14, getting an answer of 439.6 m.

Student finds the circumference of the larger garden correctly as 70𝜋𝜋 m and determines the difference between the larger and smaller gardens to be 20𝜋𝜋 m.

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c

7.G.B.4

Student does not find the areas of the gardens but adds the total circumferences of both the smaller and larger garden to get 120𝜋𝜋.

Student makes a conceptual error, such as multiplying the radius by 2 instead of squaring the radius. In this case, the areas would be 140𝜋𝜋 and 50𝜋𝜋. The total is 190𝜋𝜋 m2. OR Student makes two or more computational and/or labeling errors.

Student finds the area of both gardens correctly, in terms of 𝜋𝜋, as 1225𝜋𝜋 and 625𝜋𝜋, but does not find the total sum for both gardens. OR Student uses the area formulas correctly but makes one computational or labeling error (m2

). OR Student finds the correct areas but does not leave the answer in terms of 𝜋𝜋 and instead uses 3.14, getting an answer of 5809 m2.

Student finds the area of both gardens correctly in terms of 𝜋𝜋, as 1225𝜋𝜋 and 625𝜋𝜋, and finds the total amount of fabric needed for both gardens as 1850𝜋𝜋 m2.

4 a

7.G.B.4

Student answer is incorrect or missing. Student work shows little or no evidence of correct reasoning.

Student makes a conceptual error such as finding the circumference or the area of the semicircle but uses the diameter as 20 in doing so.

Student makes a computational error, such as dividing incorrectly.

Student correctly determines the radius of the semicircle of 20 ft. by dividing the diameter of 40 by 2.

b

7.G.B.4

Student answer is incorrect or missing. Student work shows little or no evidence of correct reasoning.

Student makes a conceptual error, such as using the wrong formulas for area or subtracting the areas as in the area of a shaded region. OR Student makes two or more computational or labeling errors.

Student makes one computational or labeling error. OR Student finds the correct area of the square, 1600 ft2, and the semicircle, 628 ft2 but does not add them to get the total area. OR Student does not use 3.14 for 𝜋𝜋 as instructed and leaves the area of the semicircle in terms of 𝜋𝜋. OR Student finds the area of the square correctly but finds the area of the entire circle, not the semicircle, while finding the answer of 2,856 ft2.

Student finds the overall area correctly as 2,228 ft2.

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5 a

7.G.B.5

Student work shows little understanding of supplementary angles.

Student makes a conceptual error such as translating the angles incorrectly, but all further work is correct. OR Student makes two or more computational errors.

Student makes one computational error in solving the equation.

Student correctly defines the variable, translates each angle into algebraic

expressions, 13𝑚𝑚 and

𝑚𝑚, writes an equation,

13𝑚𝑚 + 𝑚𝑚 = 180, solves

the equation correctly, 𝑚𝑚 = 135, and finds the measure of ∠𝐵𝐵𝐵𝐵𝐶𝐶 = 135°. Students are not limited to using equations to solve this problem. For example, they could also set up an appropriate tape diagram.

b

7.G.B.5

Student work shows little evidence of correct reasoning, such as writing an equivalent expression for 9(8𝑥𝑥 + 11) as 72𝑥𝑥 + 99, but with no further correct work shown.

Student does not write a correct equation using the answer from part (a) but solves the written equation correctly, provided it is of equal difficulty.

Student writes acorrect equation based on the answer from part (a) but makes one computational error in solving.

Student uses the answer from part (a) to find the correct value for 𝑥𝑥 = 1

2 Student may have an incorrect answer, but if the equation written is correct based on a wrong answer from part (a), and the equation is solved correctly, then full credit can be given.

6 a

7.G.B.6

Student work shows little evidence of correct reasoning.

Student makes a conceptual error such as not finding the volume and finding the surface area incorrectly. OR Student makes two or more computational and/or labeling errors.

Student uses the volume formula but makes one computational or labeling error.

Student correctly uses the volume formula for a rectangular prism to find how much water is needed to fill the pool, 2,700 ft3.

b

7.G.B.6

Student work shows little evidence of correct reasoning, such as adding or subtracting 7.48 or not using the volume from part (a).

Student makes a conceptual error such as dividing the volume by 7.48 instead of multiplying. If so, the answer would be 361 gallons.

Student knows to use the volume from part (a) and multiply by 7.48 but makes one computational error.

Student uses the answer from part (a) and multiplies it by 7.48 to find the total number of gallons needed to fill the pool. If part (a) is answered correctly, then the correct answer is 20,196 gallons.

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c

7.G.B.6

Student work shows little evidence of correct reasoning. OR Student finds the correct amount of gallons remaining, 16,830, but no further work is shown or correct.

Student makes a conceptual error such as finding the change in gallons but incorrectly uses the volume formula with gallons. For example,

16830 = 25 × 18 × ℎ ℎ = 37.4.

Student demonstrates a solid understanding but makes one computational error. OR Student correctly determines the new height, 5 ft., after the water leaked but does not find the change in the height.

Student finds the new depth of the pool after the water leaked and determines the change in the height as 1 ft. Student can solve in a number of ways, such as finding the number of gallons remaining, dividing by 7.48 to determine the new volume, setting up an equation (such as 2250 = 25 × 18 × ℎ) to determine the new height, and finally, subtracting the height from the original height to get the change. Another approach is to write an equation to find the height of the volume that was lost.

d

7.G.B.6

Student work shows little evidence of correct reasoning. OR Student finds the area of one of the sides, either 450, 108, or 150, but no further correct work is shown.

Student makes a conceptual error such as using the wrong area formulas or only finding the area of three of the surfaces. OR Student makes two or more computational or labeling errors.

Student demonstrates a solid understanding of surface area but makes one computational or labeling error. OR Student gets an answer of 1,416 ft2 by finding the surface area of all surfaces, including the top base.

Student correctly determines the surface area of the sides to be resurfaced with appropriate work shown. 25 × 18 + 2(6 × 18) +2(6 × 25) = 966 The surface area that needs to be covered is 966 ft2.

7 a–b

7.G.B.5

Student work shows little evidence of correct reasoning. OR Student writes a correct equation, but no further work or correct work is shown.

Student makes a conceptual error, writing an incorrect equation for part (a) but solves it correctly for part (b). OR Student writes a correct equation but makes a conceptual error when solving the equation, such as

2𝑚𝑚− 10 = 90 𝑚𝑚 = 50.

OR Student writes a correct equation but makes two or more computational errors.

Student writes a correct equation but makes one computational error.

Student writes and solves a correct equation,

3𝑚𝑚 + 𝑚𝑚 − 10 = 90 𝑚𝑚 = 25,

with all appropriate work shown.

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c–d

7.G.B.5

Student work shows little evidence of substituting the value of 𝑚𝑚 into the given angle measures. Instead, student assumes one angle is 25° and finds the complement for the other angle to be 65°.

Student finds one angle measure correctly with appropriate supporting work. OR Student makes two or more computational errors.

Student uses the answer from part (b), replacing the value into the given angle measures ∠𝑇𝑇𝑇𝑇𝑇𝑇 and ∠𝑇𝑇𝑇𝑇𝐿𝐿, but makes one computational error.

Student correctly uses the answer from part (b) and substitutes its value into the given angle measures to find the measures of ∠𝑇𝑇𝑇𝑇𝑇𝑇 and ∠𝑇𝑇𝑇𝑇𝐿𝐿. If student’s answer from part (b) is correct, then the measure of ∠𝑇𝑇𝑇𝑇𝑇𝑇 = 75°, and ∠𝑇𝑇𝑇𝑇𝐿𝐿 = 15°.

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Name Date

1. Gloria says the two expressions 14

(12𝑥𝑥 + 24) − 9𝑥𝑥 and −6(𝑥𝑥 + 1) are equivalent. Is she correct? Explain how you know.

2. A grocery store has advertised a sale on ice cream. Each carton of any flavor of ice cream costs $3.79.

a. If Millie buys one carton of strawberry ice cream and one carton of chocolate ice cream, write an algebraic expression that represents the total cost of buying the ice cream.

b. Write an equivalent expression for your answer in part (a).

c. Explain how the expressions are equivalent.

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3. A new park was designed to contain two circular gardens. Garden A has a diameter of 50 m, and garden B has a diameter of 70 m.

a. If the gardener wants to outline the gardens in edging, how many meters will be needed to outline the smaller garden? (Write in terms of π.)

b. How much more edging will be needed for the larger garden than the smaller one? (Write in terms of 𝜋𝜋.)

c. The gardener wishes to put down weed block fabric on the two gardens before the plants are planted in the ground. How much fabric will be needed to cover the area of both gardens? (Write in terms of 𝜋𝜋.)

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4. A play court on the school playground is shaped like a square joined by a semicircle. The perimeter around the entire play court is 182.8 ft., and 62.8 ft. of the total perimeter comes from the semicircle.

a. What is the radius of the semicircle? Use 3.14 for 𝜋𝜋.

b. The school wants to cover the play court with sports court flooring. Using 3.14 for 𝜋𝜋, how many square feet of flooring does the school need to purchase to cover the play court?

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5. Marcus drew two adjacent angles.

a. If ∠𝐴𝐴𝐵𝐵𝐵𝐵 has a measure one-third of ∠𝐵𝐵𝐵𝐵𝐶𝐶, then what is the degree measurement of ∠𝐵𝐵𝐵𝐵𝐶𝐶?

b. If the measure of ∠𝐵𝐵𝐵𝐵𝐶𝐶 is 9(8𝑥𝑥 + 11), then what is the value of 𝑥𝑥?

6. The dimensions of an above-ground, rectangular pool are 25 feet long, 18 feet wide, and 6 feet deep.

a. How much water is needed to fill the pool?

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b. If there are 7.48 gallons in 1 cubic foot, how many gallons are needed to fill the pool?

c. Assume there was a hole in the pool, and 3,366 gallons of water leaked from the pool. How many feet did the water level drop?

d. After the leak was repaired, it was necessary to lay a thin layer of concrete to protect the sides of the pool. Calculate the area to be covered to complete the job.

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7. Gary is learning about mosaics in art class. His teacher passes out small square tiles and encourages the students to cut up the tiles in various angles. Gary’s first cut tile looks like this:

a. Write an equation relating ∠𝑇𝑇𝑇𝑇𝑇𝑇 with ∠𝑇𝑇𝑇𝑇𝐿𝐿.

b. Solve for 𝑚𝑚.

c. What is the measure of ∠𝑇𝑇𝑇𝑇𝑇𝑇?

d. What is the measure of ∠𝑇𝑇𝑇𝑇𝐿𝐿?

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Documents

A Story of Ratios - SharpSchool · and rearranging the pieces so that they are lined up, alternating direction, and form a shape that resembles a rectangle. This “rectangle” has